## Area Of Bounded Regions

**1. Area Of A Curve Between Two Ordinates** Let y = f(x) be a continuous and finite function in [a,b].

**Case 1** When the curve y = f(x) lies about the x-axis

The area bounded by the curve y = f(x), the x-axis, and the ordinates x = a and x = b is given by

area = \(\int_a^b y d x\)

**Case 2** When the curve y = f(x) lies below the x-axis

The area between the curve y = f(x), the x-axis, and the ordinates x = a and x = b is given by

area = \(\int_a^b(-y) d x\).

**2. Area Of A Curve Between Two Abscissae**

**Case 1** When the curve x = f(y) lies to the right of the y-axis

The area bounded by the curve x = f(y), the y-axis, and the abscissae y = c and y = d is given by

area = \(\int_c^d x d y .\)

**Read and Learn More Class 12 Math Solutions**

**Case 2** When the curve x = f(y) lies to the left of the y-axis

The area bounded by the curve x = f(y), the y-axis, and the abscissae y = c and y = d is given by

area = \(\int_c^d(-x) d y .\)

**3. Area Between Two Curves** The area bounded by two curves y = f(x) and y = g(x), which are intersected by the ordinates x = a and x = b, is given by

area = \(\int_a^b\{f(x)-g(x)\} d x\)

## Solved Examples

**Example 1 Using integration, find the area of the region bounded by the line 2y + x = 8, the x-axis and the lines x = 2 and x = 4.**

**Solution**

The given line AB is

2y + x = 8 ⇒ y = 4 – \(\frac{1}{2}\)x. …(1)

Required area = area PLMQP

= area between the line y = 4 – \(\frac{1}{2}\)x, and the x-axis between x = 2 and x = 4

= \(\int_2^4 y_{A B} d x=\int_2^4\left(4-\frac{1}{2} x\right) d x\)

= \(\left[4 x-\frac{1}{4} x^2\right]_2^4\) = (12-7)sq units

= 5 sq units.

Hence, the required area is 5 sq units

**Example 2 Using integration, find the area of △ABC whose vertices are A(2,3), B(4,7) and C(6,2).**

**Solution**

The equation of AB is

\(\frac{y-3}{x-2}=\frac{7-3}{4-2}\) ⇒ y = 2x – 1.

The equation of BC is

\(\frac{y-7}{x-4}=\frac{2-7}{6-4}\) ⇒ y = \(-\frac{5}{2}\)x + 17.

The equation of AC is

\(\frac{y-3}{x-2}=\frac{2-3}{6-2}\) ⇒ y = \(\frac{-1}{4} x+\frac{7}{2}\).

Draw AL ⊥ OX, BM ⊥ OX and CN ⊥ OX.

Area of △ABC = (area ALMBA + area BMNCB) – (area ALNCA)

= \(\int_2^4 y_{A B} d x+\int_4^6 y_{B C} d x-\int_2^6 y_{A C} d x\)

= \(\int_2^4(2 x-1) d x+\int_4^6\left(-\frac{5}{2} x+17\right) d x-\int_2^6\left(-\frac{1}{4} x+\frac{7}{2}\right) d x\)

= \(\left[x^2-x\right]_2^4+\left[-\frac{5}{4} x^2+17 x\right]_4^6-\left[-\frac{1}{8} x^2+\frac{7}{2} x\right]_2^6\)

= \(\left[(12-2)+(57-48)-\left(\frac{33}{2}-\frac{13}{2}\right)\right]\) sq units

= 9 sq units.

Hence, the required area is 9 sq units.

**Example 3 Calculate the area bounded by the parabola y ^{2} = 4ax and its latus rectum.**

**Solution**

Let S(a,0) be the focus of the parabola y^{2} = 4ax. Then, its latus rectum LSL’ is the line parallel to the y-axis at a distance a from it.

So, its equation is x = a.

Since the equation of the parabola contains only even powers of y, it is symmetrical about the x-axis.

∴ required area

= area LOL’L

= (area LOSL) + (area SOL’S)

= 2 x (area LOSL)

= \(2 \int_0^a y d x=2 \cdot \int_0^a 2 \sqrt{a x} d x\)

= \(4 \sqrt{a} \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^a=\frac{8}{3} a^2\) sq units.

The area bounded by the parabola = \(4 \sqrt{a} \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^a=\frac{8}{3} a^2\) sq units.

**Example 4 Using integration, find the area of the region bounded by the parabola y ^{2} = 16x and the line x = 4.**

**Solution**

y^{2} = 16x is a right-handed parabola with its vertex at the origin. And, x = 4 is the line parabola to the y-axis at a distance of 4 units from it.

Also, y^{2} = 16x contains only even powers of y.

So, it is symmetrical about the x-axis.

∴ required area = area AOCA + area BOCB

= 2(area AOCA)

= \(2 \int_0^4 y d x=2 \int_0^4 \sqrt{16 x} d x\)

= \(8 \int_0^4 \sqrt{x} d x=8 \times \frac{2}{3} \times\left[x^{3 / 2}\right]_0^4\)

= \(\frac{16}{3} \times(4)^{3 / 2}=\left(\frac{16}{3} \times 8\right)=\frac{128}{3}\) sq units.

Hence, the required area is \(\frac{128}{3}\) sq units.

**Example 5 Using integration, find the area enclosed by the parabola y ^{2} = 4ax and the chord y = mx.**

**Solution**

The given equations are

y^{2} = 4ax …(1)

and y = mx …(2)

Clearly, y^{2} = 4ax is a right-handed parabola, passing through the origin.

And y = mx is a line passing through the origin.

In order to find the points of intersection of the given parabola and the given line, we solve (1) and (2) simultaneously.

Putting y = mx from (2) into (1), we get

m^{2}x^{2 }= 4ax ⇒ x(m^{2}x – 4a) = 0

⇒ x = 0 or x = \(\frac{4 a}{m^2}\).

Now, (x = 0 ⇒ y = 0) and (x = \(\frac{4 a}{m^2}\) ⇒ y = \(\frac{4 a}{m}\)).

So, the points of intersection of the given parabola and the chord are

O(0,0) and A(\(\frac{4 a}{m^2}\), \(\frac{4 a}{m}\)).

Draw AM ⊥ OX.

Required area = (area OBAMO) – (area OAMO)

= \(\int_0^{4 a / m^2}\left(y \text { for the parabola) } d x-\int_0^{4 a / m^2}(y \text { for the line }) d x\right.\)

= \(\int_0^{4 a / m^2} 2 \sqrt{a x} d x-\int_0^{4 a / m^2} m x d x\)

= \(2 \sqrt{a} \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^{4 a / m^2}-\left[\frac{m x^2}{2}\right]_0^{4 a / m^2}\)

= \(\left[\frac{4 \sqrt{a}}{3} \cdot \frac{8}{m^3} a^{3 / 2}-\frac{m}{2} \cdot \frac{16 a^2}{m^4}\right]\)

= \(\left(\frac{32 a^2}{3 m^3}-\frac{8 a^2}{m^3}\right)=\left(\frac{8 a^2}{3 m^3}\right)\)[/latex] sq units.

Hence, the required area is \(\left(\frac{8 a^2}{3 m^3}\right)\) sq units.

**Example 6 Find the area of the region {(x,y) : x ^{2} ≤ y ≤ x].**

**Solution**

Consider the equations

y = x^{2} …(1)

and y = x …(2)

Clearly, y = x^{2} is an upward parabola and y = x is a line passing through (0,0).

Solving (1) and (2) simultaneously, we get

x^{2} = x ⇒ x^{2} – x = 0 ⇒ x(x-1) = 0

⇒ x = 0 or x = 1.

From (2), (x = 0 ⇒ y = 0) and (x = 1 ⇒ y = 1).

So, the points of intersection of (1) and (2) are O(0,0) and A(1,1).

Draw AM ⊥ OX.

Required area = shaded area

= (area OMAO) – (area OMACO)

= \(\int_0^1(y \text { for line }) d x-\int_0^1(y \text { for parabola }) d x\)

= \(\int_0^1 x d x-\int_0^1 x^2 d x=\left[\frac{x^2}{2}\right]_0^1-\left[\frac{x^3}{3}\right]_0^1=\left[\frac{1}{2}-\frac{1}{3}\right]=\frac{1}{6}\) sq unit.

Hence, the required area is \(\frac{1}{6}\) sq unit.

**Example 7 Find the area of the region bonded by the curve x ^{2} = 4y and the line x = 4y – 2.**

**Solution**

The given curve is x^{2} = 4y. …(1)

The given line is x = 4y – 2 …(2)

Putting 4y = (x+2) from (2) into (1), we get

x^{2} = (x+2) ⇔ (x^{2}-x-2) = 0

⇔ (x-2)(x+1) = 0

⇔ x = 2 or x = -1.

Putting x = 2 in (1), we get y = 1.

Putting x = -1 in (1), we get y = \(\frac{1}{4}\).

Thus, the points of intersection of the given curve (1) and the line (2) are A(-1, \(\frac{1}{4}\)) and B(2,1).

Draw AL and BM as perpendicular on the x-axis.

∴ required area = area AOBA

= (area ALMBA) – (area AOBMLA)

= \(\int_{-1}^2\left(\frac{x+2}{4}\right) d x-\int_{-1}^2 \frac{x^2}{4} d x\)

= \(\int_{-1}^2\left\{\frac{(x+2)}{4}-\frac{x^2}{4}\right\} d x=\frac{1}{4}\left[\frac{x^2}{2}+2 x-\frac{x^3}{3}\right]_{-1}^2\)

= \(\frac{1}{4}\left[\left(2+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)\right]=\frac{9}{8}\) sq units.

Hence, the required area is \(\frac{9}{8}\) sq units.

**Example 8 Find the area bounded by the circle x ^{2} + y^{2} = 16 and the line y = x in the first quadrant.**

**Solution**

The given circle is x^{2} + y^{2} = 16 …(1)

The given line is y = x

Putting y = x from (2) into (1), we get

2x^{2} = 16 ⇔ x^{2} = 8 ⇔ x = 2√2 [∵ x is +ve in the first quad.].

Thus, the point of intersection of (1) and (2) in the first quadrant is A(2√2, 2√2).

Draw AL perpendicular on the x-axis.

∴ required area = (area OLA) + (area LBAL)

= \(\int_0^{2 \sqrt{2}} x d x+\int_{2 \sqrt{2}}^4 \sqrt{16-x^2} d x\)

= \(\left[\frac{x^2}{2}\right]_0^{2 \sqrt{2}}+\left[\frac{x \sqrt{16-x^2}}{2}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{2 \sqrt{2}}^4\)

= \(\frac{1}{2}\left[(2 \sqrt{2})^2-0\right]+\left[\left(0+8 \sin ^{-1} 1\right)-\left(4+8 \sin ^{-1} \frac{1}{\sqrt{2}}\right]\right.\)

= \(\left[4+\left(8 \times \frac{\pi}{2}\right)-4-\left(8 \times \frac{\pi}{4}\right)\right]=(2 \pi) sq units.\)

The area bounded by the circle = \(\left[4+\left(8 \times \frac{\pi}{2}\right)-4-\left(8 \times \frac{\pi}{4}\right)\right]=(2 \pi) sq units.\)

**Example 9 Using integration, find the area of △ABC, whose vertices are A(2,0), B(4,5) and C(6,3).**

**Solution**

The equation of side AB is

\(\frac{y-0}{x-2}=\frac{(5-0)}{(4-2)}\) ⇒ y = \(\frac{5}{2}\)(x-2) …(1)

The equation of side BC is

\(\frac{y-0}{x-2}=\frac{(3-0)}{(6-4)}\) ⇒ y = -x + 9 …(2)

The equation of side AC is

\(\frac{y-0}{x-2}=\frac{(3-0)}{(6-2)}\) ⇒ y = \(\frac{3}{4}\) (x-2) …(3)

Draw perpendiculars BL and CM on the x-axis.

∴ area of △ABC

= ar(△ ALB) + ar(trap. BLMC) – ar(△ AMC)

= \(\int_2^4 y_{A B} d x+\int_4^6 y_{B C} d x-\int_2^6 y_{A C} d x\)

= \(\frac{5}{2} \int_2^4(x-2) d x+\int_4^6(9-x) d x-\frac{3}{4} \int_2^6(x-2) d x\)

= \(\frac{5}{2}\left[\frac{x^2}{2}-2 x\right]_2^4+\left[9 x-\frac{x^2}{2}\right]_4^6-\frac{3}{4} \cdot\left[\frac{x^2}{2}-2 x\right]_2^6\)

= \(\frac{5}{2}[0-(-2)]+(36-28)-\frac{3}{4}[6-(-2)]\)

= (5+8-6) sq units = 7 sq units.

The area of △ABC = 7 sq units.

**Example 10 Find the area cut off from the parabola 4y = 3x ^{2} by the straight line 3x – 2y + 12 = 0.**

**Solution**

Clearly, 4y = 3x^{2} is an upward parabola with its vertex at (0,0).

And, 3x – 2y + 12 = 0 is a line.

The given equations are

4y = 3x^{2} …(1)

and 3x – 2y + 12 = 0 …(2)

The points of intersection of the given parabola and the given line will be obtained by solving (1) and (2) simultaneously.

Putting y = \(\frac{3}{4}\)x^{2} from (1) in (2),

we get

\(3 x-\frac{3}{2} x^2+12=0\) ⇒ x^{2} – 2x – 8 = 0

⇒ x^{2} – 4x + 2x – 8 = 0

⇒ x(x-4) + 2(x-4) = 0

⇒ (x-4)(x+2) = 0

⇒ x = -2 or x = 4.

Now, (x = -2 ⇒ y = 3) and (x = 4 ⇒ y = 12).

So, the points of intersection of (1) and (2) are A(-2,3) and B(4,12).

Draw AL ⊥ OX’ and BM ⊥ OX.

Required area = (area ALMBA) – (area ALOMBOA)

= \(\int_{-2}^4(y \text { of the line }) d x-\int_{-2}^4(y \text { of the parabola }) d x\)

= \(\int_{-2}^4 \frac{(3 x+12)}{2} d x-\int_{-2}^4 \frac{3 x^2}{4} d x=\left[\frac{3 x^2}{4}+6 x\right]_{-2}^4-\left[\frac{x^3}{4}\right]_{-2}^4\)

= (45 – 18) = 27 sq units.

Hence, the requires area is 27 sq units.

**Example 11 Find the area bounded by the line y = x, the x-axis and the ordinates x = -1, x = 2.**

**Solution**

We know that y = x is the line passing through the origin and making an angle of 45° with x-axis, as shown in the given figure.

Now, we have to find the area of the shaded region.

Required area

= (area ODBO) + (area OACO)

= \(\int_0^2 y d x+\int_{-1}^0(-y) d x\) [∵ area OACO is below the x-axis]

= \(\int_0^2 x d x+\int_{-1}^0(-x) d x\)

= \(\left[\frac{x^2}{2}\right]_0^2+\left[\frac{-x^2}{2}\right]_{-1}^0=\left[2+\frac{1}{2}\right]=\frac{5}{2}\) sq units.

Hence, the required area is \(\frac{5}{2}\) sq units.

**Example 12 Find by integration, the area of the region bounded by the curve y = 2x – x ^{2} and the x-axis.**

**Solution**

The given curve is y = 2x-x^{2 } …(1)

Now, y = 2x – x^{2} ⇒ (x^{2}-x^{2}+1) = (-y+1)

⇒ (x-1)^{2} = -1(y-1)

⇒ x^{2} = -y,

where (x-1) = X and (y-1) = Y.

Clearly, x^{2} = -Y is a downward parabola with its vertex at (X=0, Y=0).

Now,

X = 0, Y = 0 ⇒ x – 1 = 0 and y – 1 = 0

⇒ x = 1 and y = 1.

Thus, the vertex of the parabola is A(1,1).

Also, y = 0 ⇒ 2x-x^{2} = 0 ⇒ x(2-x) = 0 ⇒ x = 0 or x = 2.

Thus, the curve cuts the x-axis at O(0,0) and B(2,0).

The rough sketch of the curve can now be drawn, as shown in the given figure.

∴ required area = \(\int_0^2 y d x\)

= \(\int_0^2\left(2 x-x^2\right) d x=\left[x^2-\frac{x^3}{3}\right]_0^2\)

= \(\left(4-\frac{8}{3}\right)=\frac{4}{3}\)

Hence, the required area is \(\frac{4}{3}\) sq units.

**Example 13 Using integration, find the area of the region bounded by the ellipse** \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

**Solution**

The given equation contains only even powers of y.

So, the curve is symmetrical about the x-axis.

Also, the given equation contains only even powers of x.

So, the curve is symmetrical about the y-axis.

A rough sketch of the ellipse can be drawn, as shown in the figure.

Now, \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ⇒ y = \frac{b}{a} \cdot \sqrt{a^2-x^2} .\)

Area of the given ellipse

= 4 x (area OBCO)

= \(4 \times \int_0^a y d x=4 \times \int_0^\pi \frac{b}{a} \cdot \sqrt{a^2-x^2} d x\)

= \(\frac{4 b}{a} \cdot \int_0^{\pi / 2} a^2 \cos ^2 \theta d \theta\) [putting x = a sin θ so that dx = a cos θ dθ]

= \((4 a b) \cdot \int_0^{\pi / 2} \frac{(1+\cos 2 \theta)}{2} d \theta=(2 a b) \cdot\left[\theta+\frac{\sin 2 \theta}{2}\right]_0^{\pi / 2}\)

= (πab) sq units.

Hence, the area of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is (πab) sq units.

**Example 14 By using integration, prove that the area of a circle of radius r units is πr ^{2} square units.**

**Solution**

The equation of a circle of radius r units with its centre at the origin is

x^{2} + y^{2} = r^{2} …(1)

This equation contains only even powers of y.

So, the curve is symmetrical about the x-axis.

Also, the above equation contains only even powers of x.

So, the curve is symmetrical about the y-axis.

Now, x^{2} + y^{2} = r^{2} ⇒ y = \(\sqrt{r^2-x^2}\).

Area of the circle = 4 x (area OABO)

= \(4 \times \int_0^r y d x=4 \times \int_0^r \sqrt{r^2-x^2} d x\)

= \(4 \times \int_0^{\pi / 2} r^2 \cos ^2 \theta d \theta\) [putting x = r sin θ]

= \(4 r^2 \cdot \int_0^{\pi / 2} \frac{(1+\cos 2 \theta)}{2} d \theta=2 r^2 \cdot\left[\theta+\frac{\sin 2 \theta}{2}\right]_0^{\pi / 2}\)

= (πr^{2}) sq units.

Hence, the required area is (πr^{2}) sq units.

**Example 15 Find the area of the smaller region bounded by the ellipse** \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) **and the straight line** \(\frac{x}{a}+\frac{y}{b}=1 .\)

**Solution**

The ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and the line \(\frac{x}{a}+\frac{y}{b}=1\) can be drawn, as shown in the given figure.

Then, we have to find the area of the shaded region.

Required area = {area between \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and the x-axis from x = 0 to x = a} – {area between the line \(\frac{x}{a}+\frac{y}{b}=1\) and the x-axis from x = 0 to x = a}

= \(\int_0^a(y \text { of the ellipse }) d x-\int_0^a(y \text { of the line }) d x\)

= \(\int_0^a \frac{b}{a} \cdot \sqrt{a^2-x^2} d x-\int_0^a \frac{b(a-x)}{a} d x\)

= \(\frac{b}{a} \cdot\left[\frac{x \sqrt{a^2-x^2}}{2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a-\frac{b}{a} \cdot\left[a x-\frac{x^2}{2}\right]_0^a\)

= \(\frac{a b}{2}\left(\sin ^{-1} 1-\sin ^{-1} 0\right)-\left(a b-\frac{a b}{2}\right)=\left(\frac{\pi a b}{4}-\frac{a b}{2}\right)\) sq units.

Hence, the required area is \(\left(\frac{\pi a b}{4}-\frac{a b}{2}\right)\) sq units.

**Example 16 Find the area of the region bounded by the parabola x ^{2} = y and y^{2} = x.**

**Solution**

Clearly, x^{2} = y is an upward parabola with its vertex at (0,0) and y^{2} = x is a right-handed parabola with its vertex also at (0,0).

The shaded region shows the area bounded by these parabolas.

The given equations are

x^{2} = y …(1)

and y^{2} = x …(2)

Using (1) in (2), we get

x^{4} = x ⇒ x(x^{3}-1) = 0 ⇒ x = 0 or x = 1.

Also, (x = 0 ⇒ y = 0) and (x = 1 ⇒ y = 1).

So, the given curves intersect at O(0,0) and A(1,1).

Draw AL ⊥ OX.

Required area = (area OLABO) – (area OLACO)

= {area bounded by y^{2} = x from x = 0 to x = 1} – {area bounded by x^{2} = y from x = 0 to x = 1}

= \(\int_0^1 \sqrt{x} d x-\int_0^1 x^2 d x\)

= \(\left[\frac{2}{3} x^{3 / 2}\right]_0^1-\left[\frac{x^3}{3}\right]_0^1=\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{1}{3}\) unit.

Hence, the required area is \(\frac{1}{3}\) unit.

**Example 17 Find the area of the region included between the parabolas y ^{2} = 4ax and x^{2} = 4ay, where a > 0.**

**Solution**

The given parabolas are

y^{2} = 4ax …(1)

and x^{2} = 4ay …(2)

In order to find the points of intersection of the given curves, we solve (1) and (2) simultaneously.

Putting x = \(\frac{y^2}{4 a}\) from (1) in (2), we get

\(\frac{y^4}{16 a^2}=4 a y\) ⇒ y4 – 64a^{3}y = 0

⇒ y(y^{3} – 64a^{3}) = 0

⇒ y = 0 or y = 4a.

Now, (y = 0 ⇒ x = 0) and (y = 4a ⇒ x = \(\frac{16 a^2}{4 a}\) = 4a).

Thus, the points of intersection of the two parabolas are O(0,0) and A(4a, 4a).

Draw AD ⊥ OX. Then, point D is (4a,0).

Required area = area OCABO

= (area OBADO) – (area OCADO)

= \(\int_0^{4 \pi} y d x \text { for }\left(y^2=4 a x\right)-\int_0^{4 a} y d x \text { for }\left(x^2=4 a y\right)\)

= \(\int_0^{4 a} 2 \sqrt{a x} d x-\int_0^{4 a} \frac{x^2}{4 a} d x\)

= \(\left[2 \sqrt{a} \cdot \frac{2}{3} \cdot x^{3 / 2}\right]_0^{4 t}-\frac{1}{4 a}\left[\frac{x^3}{3}\right]_0^{4 \pi}\)

= \(\left[\frac{4 \sqrt{a}}{3} \cdot(4 a)^{3 / 2}-\frac{1}{12 a} \times 64 a^3\right]\)

= \(\left(\frac{32 a^2}{3}-\frac{16 a^2}{3}\right)=\left(\frac{16 a^2}{3}\right)\) sq units.

Hence, the required area is \(\left(\frac{16 a^2}{3}\right)\) sq units.

**Example 18 Using integration, find the area of the region common to the circle x ^{2} + y^{2} = 16 and the parabola y^{2} = 6x.**

**Solution**

x^{2} + y^{2} = 16 is a circle with its centre at (0,0) and radius = 4 units. And, y^{2} = 6x is a right-handed parabola with its vertex at (0,0).

Now, we have to find the area of the shaded region.

The given equations are

x^{2} + y^{2} = 16 …(1)

and y^{2} = 6x …(2)

Using (2) in (1), we get

x^{2} + 6x – 16 = 0 ⇒ (x+8)(x-2) = 0

⇒ x = -8 or x = 2.

Now, x = -8 ⇒ y^{2} = -48

⇒ y is imaginary.

And, x = 2 ⇒ y^{2} = (6 x 2) = 12

⇒ y = ±2√3.

Thus, the points of intersection of the given curves are A(2, 2√3) and B(2, -2√3).

Since each of the given equations contains only even powers of y, each one is symmetrical about the x-axis.

∴ required area = 2(area OCDAO)

= 2(area OCAO + area CDAC)

= \(2\left[\int_0^2 y d x \text { for curve (ii) }+\int_2^4 y d x \text { for curve (i) }\right]\)

= \(2\left[\int_0^2 \sqrt{6 x} d x+\int_2^4 \sqrt{16-x^2} d x\right]\)

= \(2\left\{\left[\frac{2 \sqrt{6}}{3} x^{3 / 2}\right]_0^2+\left[x \sqrt{\frac{16-x^2}{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_2^4\right\}\)

= \(\left\{\left(\frac{2 \sqrt{6}}{3} \cdot 2^{3 / 2}-0\right)+8 \sin ^{-1} 1-\left(2 \sqrt{3}+8 \sin ^{-1} \frac{1}{2}\right)\right\}\)

= \(2\left(\frac{8 \sqrt{3}}{3}+4 \pi-2 \sqrt{3}-\frac{4 \pi}{3}\right)=2\left(\frac{2 \sqrt{3}}{3}+\frac{8 \pi}{3}\right)\)

= \(\frac{4}{3}(\sqrt{3}+4 \pi)\) sq units.

Hence, the required area is \(\frac{4}{3}(\sqrt{3}+4 \pi)\) sq units.

**Example 19 Using integration, find the area of the region enclosed between the circles x ^{2} + y^{2} = 4 and (x-2)^{2} + y^{2} = 4.**

**Solution**

x^{2} + y^{2} = 4 is a circle with its centre at O(0,0) and radius = 2 units.

And, (x-2)^{2} + y^{2} = 4 is a circle with its centre at C(2,0) and radius = 2 units.

The given circles are

x^{2} + y^{2} = 4 …(1)

and (x-2)^{2} + y^{2} = 4 …(2)

Eliminating y from (1) and (2), we get

4 – x^{2} = 4 – (x-2)^{2} ⇒ 4x = 4 ⇒ x = 1.

Putting x = 1 in (1), we get y^{2} = 3 ⇒ y = ±√3.

Thus, the points of intersection of the two circles are A(1, √3) and B(1, -√3).

Both the circles are symmetrical about the x-axis.

Required area = 2(area AOCA)

= 2(area AODA + area CADC)

= \(2 \int_0^1 y d x \text { for circle (ii) }+2 \int_1^2 y d x \text { for circle (i) }\)

= \(2 \int_0^1 \sqrt{4-(x-2)^2} d x+2 \int_1^2 \sqrt{4-x^2} d x\)

= \(\left[\frac{(x-2) \sqrt{4-(x-2)^2}}{2}+\frac{4}{2} \cdot \sin ^{-1} \frac{(x-2)}{2}\right]_0^1+\left[\frac{x \sqrt{4-x^2}}{2}+\frac{4}{2} \cdot \sin ^{-1} \frac{x}{2}\right]_1^2\)

= \(2\left[\left\{\frac{-\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{-1}{2}\right)\right\}-\left\{0+2 \sin ^{-1}(-1)\right\}\right.\left.+\left\{2 \sin ^{-1}(1)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\frac{1}{2}\right)\right\}\right]\)

= \(2\left\{\frac{-\sqrt{3}}{2}+2\left(-\frac{\pi}{6}\right)-2\left(-\frac{\pi}{2}\right)+\left(2 \cdot \frac{\pi}{2}-\frac{\sqrt{3}}{2}-2 \times \frac{\pi}{6}\right)\right\}\)

= \(2\left(\frac{4 \pi}{3}-\sqrt{3}\right)\) sq units.

Hence, the required area is \(2\left(\frac{4 \pi}{3}-\sqrt{3}\right)\) sq units.

**Example 20 Using integration, find the area of the region {(x,y) : y ^{2} ≤ 4x, 4x^{2} + 4y^{2} ≤ 9}.**

**Solution**

Clearly, we have to find the area enclosed between the curves y^{2} = 4x and 4x^{2} + 4y^{2} = 9.

The curve y^{2} = 4x is a right-handed parabola with its vertex at (0,0).

4x^{2} + 4y^{2} = 9 ⇒ x^{2} + y^{2} = \(\left(\frac{3}{2}\right)^2\), which represents a circle with its centre at O(0,0) and radius equal to (3/2) units.

The shaded region shows the required area.

Now, y^{2} = 4x …(1)

and 4x^{2} + 4y^{2} = 9 …(2)

Putting the value of y^{2} from (1) into (2), we get

4x^{2} + 16x – 9 = 0

⇒ 4x^{2} + 18x – 2x – 9 = 0

⇒ 2x(2x+9) – (2x+9) = 0

⇒ (2x+9)(2x-1) = 0

⇒ x = \(-\frac{9}{2}\) or x = \(\frac{1}{2}\).

Putting x = \(-\frac{9}{2}\) in (1), we get

y^{2} = -18 ⇒ y is imaginary.

Putting x = \(\frac{1}{2}\) in (1), we get y^{2} = 2 ⇒ y = ±√2.

So, the given curves intersect at the points A(\(\frac{1}{2}\), √2) and B(\(\frac{1}{2}\), -√2).

Since the equation of each of the given curves contains only even powers of y, each curve is symmetrical about the x-axis.

Required area = 2(area ODCAO)

= 2(area ODAO + area DCAD)

= \(2\left\{\int_0^{1 / 2}(y \text { of the parabola }) d x+\int_{1 / 2}^{3 / 2}(y \text { of the circle }) d x\right\}\)

= \(2\left\{\int_0^{1 / 2} 2 \sqrt{x} d x+\int_{1 / 2}^{3 / 2} \sqrt{\frac{9}{4}-x^2} d x\right\}\)

= \(2\left[\left[\frac{4}{3} x^{3 / 2}\right]_0^{1 / 2}+\left[\frac{x \sqrt{\frac{9}{4}-x^2}}{2}+\frac{9}{8} \sin ^{-1}\left(\frac{x}{3 / 2}\right)\right]_{1 / 2}^{3 / 2}\right\}\)

= \(\frac{4}{3 \sqrt{2}}+\left(\frac{9}{4} \sin ^{-1} 1\right)-\left(\frac{\sqrt{2}}{2}+\frac{9}{4} \sin ^{-1} \frac{1}{3}\right)\)

= \(\left(\frac{2 \sqrt{2}}{3}-\frac{\sqrt{2}}{2}\right)+\frac{9}{4}\left(\sin ^{-1} 1-\sin ^{-1} \frac{1}{3}\right)\)

= \(\left\{\frac{\sqrt{2}}{6}+\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}\right)\right\}\) sq units

= \(\left\{\frac{\sqrt{2}}{6}+\frac{9}{4} \cos ^{-1}\left(\frac{1}{3}\right)\right\}\) sq units.

Hence, the required area is \(\left\{\frac{\sqrt{2}}{6}+\frac{9}{4} \cos ^{-1}\left(\frac{1}{3}\right)\right\}\) sq units.

**Example 21 Find the area of the region {(x,y): x ^{2} + y^{2} ≤ 1 ≤ x + y}.**

**Solution**

Let R = {(x,y) : x^{2} + y^{2} ≤ 1 ≤ x+y}

= {(x,y) : x^{2} + y^{2} ≤ 1} ∩ {(x,y) : x + y ≥ 1}

= R_{1} ∩ R_{2}.

Clearly, R_{1} is the interior of the circle x^{2} + y^{2} = 1 with its centre at O(0,0) and radius = 1 unit.

And, R_{2 }is the region lying above the line

Consider the equations

x^{2} + y^{2} = 1 …(1)

and x + y = 1 …(2)

Putting y = (1-x) from (2) in (1), we get

x^{2} + (1-x)^{2 }= 1 ⇒ 2x^{2} – 2x = 0

⇒ 2x(x-1) = 0

⇒ x = 0 or x = 1.

Now, (x = 0 ⇒ y = 1) and (x = 1 ⇒ y = 0).

Thus, the points of intersection of (1) and (2) are A(0,1) and B(1,0).

So, the required area is the shaded region.

Required area = area BCAB

= (area AOBCA) – (area OBAO)

= \(\int_0^1 \sqrt{1-x^2} d x-\int_0^1(1-x) d x\)

= \(\left[\frac{1}{2} \sin ^{-1} x+\frac{x}{2} \sqrt{1-x^2}\right]_0^1-\left[x-\frac{x^2}{2}\right]_0^1\)

= \(\left(\frac{1}{2} \sin ^{-1} 1\right)-\frac{1}{2}=\left(\frac{1}{2} \times \frac{\pi}{2}\right)-\frac{1}{2}=\left(\frac{\pi}{4}-\frac{1}{2}\right)\) sq units.

Hence, the required area is \(\left(\frac{\pi}{4}-\frac{1}{2}\right)\) sq units.

**Example 22 Find the area of the region {(x,y) : x ^{2} + y^{2} ≤ 2ax, y^{2} ≥ ax, x ≥ 0, y ≥ 0}.**

**Solution**

Clearly, we have to find the area of the region lying in the first quadrant (x ≥ 0, y ≥ 0), included between the circle x^{2} + y^{2} = 2ax and the parabola y^{2} = ax.

Thus, the equations of the curves are

x^{2} + y^{2} = 2ax …(1)

and y^{2} = ax …(2)

Now, clearly x^{2} + y^{2} = 2ax is a circle with its centre B(a,0) and radius = a units.

And, y^{2} = ax is a parabola with O(0,0) as its vertex and the x-axis as its axis.

We can draw the figure, as shown.

Their points of intersection may be obtained by solving (1) and (2) and keeping in view that x ≥ 0 and y ≥ 0.

Using (2) in (1), we get

x^{2} – ax = 0 ⇒ x(x-a) = 0

⇒ x = 0 or x = a.

Now, (x = 0 ⇒ y = 0) and (x = a ⇒ y = a).

Thus, the two curves intersect at O(0,0) and A(a,a).

∴ required area = \(\int_0^a \sqrt{2 a x-x^2} d x-\int_0^a \sqrt{a x} d x\)

= \(\int_0^a \sqrt{a^2-(x-a)^2} d x-\sqrt{a} \cdot \int_0^a \sqrt{x} d x\)

= \(\left[\frac{(x-a) \sqrt{a^2-(x-a)^2}}{2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)\right]_0^a-\sqrt{a}\left[\frac{2}{3} x^{3 / 2}\right]_0^a\)

= \(\left\{\frac{a^2}{2} \sin ^{-1}(0)-\frac{a^2}{2} \sin ^{-1}(-1)-\frac{2}{3} a^2\right\}\)

= \(\left(\frac{\pi a^2}{4}-\frac{2}{3} a^2\right)\) sq units.

Hence, the required area is \(\left(\frac{\pi a^2}{4}-\frac{2}{3} a^2\right)\) sq units.

**Example 23 Find the area of the region {(x,y) : x ^{2} ≤ y ≤ |x|}.**

**Solution**

Consider the equations

x^{2} = y …(1)

and y = |x| …(2)

Clearly, x^{2} = y represents an upward parabola with its vertex at O(0,0). All the points inside this parabola represent x^{2} ≤ y.

Also, y = |x| = \(\left\{\begin{array}{r}

x, \text { when } x \geq 0 \\

-x, \text { when } x<0

\end{array}\right.\)

The lines OA and OB, each equally inclined to the axes, represent y = |x|. All the points below the lines OA and OB, and above the x-axis represent y ≤ |x|.

Thus, the shaded portion is the required region.

In each of the given equations, the equation remains unchanged when x is replaced by -x.

So, each of the given curves is symmetrical about the y-axis.

∴ required area = 2(area OEAO).

In this region, we have

x^{2} = y …(3)

and y = x …(4)

Using (4) in (3), we get x^{2} = x ⇒ x(x-1) = 0 ⇒ x = 0 or x = 1.

Now, (x = 0 ⇒ y = 0) and (x = 1 ⇒ y = 1).

Thus, the line y = x and the curve x^{2} = y intersect at O(0,0) and A(1,1). Draw AD ⊥ OX and BC ⊥ OX’.

∴ required area = 2(area OEAO) = 2[(area ODAO) – (area ODAEO)]

= 2[(area between y = x and the x-axis from x = 0 to x = 1) – (area between x2 = y and the x-axis from x = 0 to x = 1)]

= \(2\left(\int_0^1 y d x \text { for the line } O A-\int_0^1 y d x \text { for the curve } O E A\right)\)

= \(2\left[\int_0^1 x d x-\int_0^1 x^2 d x\right]=2\left\{\left[\frac{x^2}{2}\right]_0^1-\left[\frac{x^3}{3}\right]_0^1\right\}=2\left[\frac{1}{2}-\frac{1}{3}\right]\)

= \(\left(2 \times \frac{1}{6}\right)=\frac{1}{3}\) sq unit.

Hence, the required area is \(\frac{1}{3}\) sq unit.

**Example 24 Find the area bounded by the line y = x and the curve y = x ^{3}.**

**Solution**

The given equations are

y = x …(1)

and y = x^{3} …(2)

Using (1) in (2), we get

x – x^{3} = 0 ⇒ x(1-x^{2}) = 0 ⇒ x(1-x)(1+x) = 0

⇒ x = 0 or x = 1 or x = -1.

Also, (x = 0 ⇒ y = 0), (x = 1 ⇒ y = 1) and (x = -1 ⇒ y = -1).

So, the given curve and the line intersect at the points O(0.0), A(1,1) and B(-1, -1).

Now, y = x is a line passing through the origin and making an angle of 45° with the x-axis. Thus, the line y = x can be drawn.

For the curve y = x^{3} some values for x and the corresponding values of y are given below:

Plotting the points (1,-1), (\(-\frac{1}{2}\), \(-\frac{1}{8}\)), (0,0), (\(\frac{1}{2}\), \(\frac{1}{8}\)) and (1,1) and joining them, we get a rough sketch of y = x^{3}, as shown in the given figure.

Required area = (area ACOA) + (area ODBO)

= (area OALO) – (area OCALO) + (area OBMO) – (area ODBMO)

= \(\int_0^1\{y \text { for (i) }\} d x-\int_0^1\{y \text { for (ii) }\} d x+\int_{-1}^0\{(-y) \text { for (i) }\} d x-\int_{-1}^0\{(-y) \text { for (ii) }\} d x\)

= \(\int_0^1 x d x-\int_0^1 x^3 d x+\int_{-1}^0-x d x-\int_{-1}^0-x^3 d x\)

= \(\left[\frac{x^2}{2}\right]_0^1-\left[\frac{x^4}{4}\right]_0^1+\left[\frac{-x^2}{2}\right]_{-1}^0+\left[\frac{x^4}{4}\right]_{-1}^0\)

= \(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{2}\) sq unit.

Hence, the required area is 0.5 sq unit.

**Example 25 Find the area bounded by the curve y = sin x between x = 0 and x = 2π.**

**Solution**

The given curve is y = sin x.

Some values of x and the corresponding values of y are given below.

Taking a fixed unit (distance) for π along the x-axis, we can plot the points (0,0), (\(\frac{\pi}{6}\), \(\frac{1}{2}\)), (\(\frac{\pi}{2}\), 1), (π, 0), (\(\frac{3 \pi}{2}\), -1) and (2π, 0).

Join these points freehand to obtain a rough sketch of the given curve.

Required area = (area OABO) + (area BCDB)

= \(\int_0^\pi y d x+\int_\pi^{2 \pi}(-y) d x\) [∵ area BCDB is below the x-axis]

= \(\int_0^\pi \sin x d x-\int_\pi^{2 \pi} \sin x d x\)

= \([-\cos x]_0^\pi-[-\cos x]_\pi^{2 \pi}=(2+2)=4\) sq units.

Hence, the required area is 4 sq units.

**Example 26 Find the area of the region bounded by the curve y = x ^{2}+2, and the lines y = x, x = 0 and x = 3.**

**Solution**

y = x^{2} + 2 ⇒ x^{2} = (y-2).

Clearly, x^{2} = (y-2) represents an upward parabola with its vertex at A(0,2).

Also, y = x represents the straight line, making an angle of 45° with the positive direction of the x-axis.

And, x = 0 is the y-axis, while x = 3 represents a line parallel to the y-axis at a distance of 3 units from it.

Thus, the shaded region in the given figure is the required area.

∴ required area = (area ODCAO) – (area ODBO)

= \(\int_0^3\left(x^2+2\right) d x-\int_0^3 x d x\)

= \(\left[\frac{x^3}{3}+2 x\right]_0^3-\left[\frac{x^2}{2}\right]_0^3=\left(15-\frac{9}{2}\right)=\frac{21}{2}\) sq units.

Hence, the required area is \(\frac{21}{2}\) sq units.

**Example 27 Find the area of the region {(x,y) : 0 ≤ y ≤ (x ^{2}+1), 0 ≤ y ≤ (x+1), 0 ≤ x ≤ 2}.**

**Solution**

Let R = {(x,y) : 0 ≤ y ≤ (x^{2}+1), 0 ≤ y ≤ (x+1), 0 ≤ x ≤ 2}

= {(x,y) : 0 ≤ y ≤ (x^{2}+1)} ∩ {(x,y) : 0 ≤ y ≤ (x+1)} ∩ {(x,y) : 0 ≤ x ≤ 2}

= R_{1} ∩ R_{2} ∩ R_{3}.

Clearly, R_{1} is the region consisting of the right-hand side of the y-axis, lying below the parabola y = x^{2} + 1.

Also, R_{2} is the region consisting of the right-hand side of the y-axis, lying below the line y = (x+1).

And, R_{3} is the region above the x-axis, lying between the ordinates x = 0 and x = 2.

Thus, R_{1 }∩ R_{2} ∩ R_{3} is the shaded region.

We have, y = x^{2} + 1 and y = x + 1

⇒ x^{2} + 1 = x + 1 ⇒ x(x-1) = 0 ⇒ x = 0 or x = 1.

Now, (x = 0 ⇒ y = 1) and (x = 1 ⇒ y = 2).

Thus, the parabola y = (x^{2}+1) and the line y = x + 1 intersect at the points A(0,1) and C(1,2).

∴ required area = area of the shaded region

= (area ODCBA) + (area CDFEC)

= \(\int_0^1(y \text { of the parabola }) d x+\int_1^2(y \text { of the line }) d x\)

= \(\int_0^1\left(x^2+1\right) d x+\int_1^2(x+1) d x\)

= \(\left[\frac{x^3}{3}+x\right]_0^1+\left[\frac{x^2}{2}+x\right]_1^2\)

= \(\left(\frac{1}{3}+1\right)+\left(4-\frac{3}{2}\right)=\frac{23}{6}\) sq units.

Hence, the required area is \(\frac{23}{6}\) sq units.

**Example 28 Find the area of the region bounded by the curve y ^{2} = 2y – x and the y-axis.**

**Solution**

y^{2} = 2y – x ⇒ y^{2} – 2y = -x

⇒ y^{2} – 2y + 1 = -x + 1

⇒ (y-1)^{2} = -(x-1)

⇒ y^{2} = -X,

Where y – 1 = Y and (x-1) = X.

This is a left-handed parabola with vertex at (X = 0, Y = 0).

X = 0, Y = 0 ⇒ -x + 1 = 0 and y – 1 = 0

⇒ x = 1 and y = 1.

Thus, the vertex of the given parabola is A(1,1).

Also, x = 0 ⇒ y^{2 }– 2y = 0 ⇒ y(y-2) = 0 ⇒ y = 0 or y = 2.

Thus, the curve meets the y-axis at O(0,0) and B(0,2).

A rough sketch of the curve can be drawn, as shown in the figure.

∴ required area = \(\int_0^2 x d y=\int_0^2\left(2 y-y^2\right) d y\)

= \(\left[y^2-\frac{y^3}{3}\right]_0^2=\left(4-\frac{8}{3}\right)=\frac{4}{3}\) sq units.

Hence, the required area is \(\frac{4}{3}\) sq units.