# NCERT Solutions For Class 12 Maths Chapter 8 Exercise 8.1

## Chapter 8 Ex.8.1 Question 1

Find the area of the region bounded by the curve $${y^2} = x$$ and the lines $$x = 1,x = 4$$ and the x-axis in the first quadrant.

### Solution

\begin{align}ar\left( {ABCD} \right) &= \int_1^4 {ydx} \\ &= \int_1^4 {\sqrt x dx} \\& = \left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_1^4\\ &= \frac{2}{3}\left[ {{{\left( 4 \right)}^{\frac{3}{2}}} - {{\left( 1 \right)}^{\frac{3}{2}}}} \right]\\ &= \frac{2}{3}\left[ {8 - 1} \right]\\ &= \frac{{14}}{3}\end{align}

## Chapter 8 Ex.8.1 Question 2

Find the area of the region bounded by $${y^2} - 9x,x = 2,x = 4$$ and the x-axis in the first quadrant.

### Solution

\begin{align}ar\left( {ABCD} \right) &= \int_2^4 {ydx} \\ &= \int_2^4 {3\sqrt x dx} \\ & = 3\left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_2^4\\& = 2\left[ {{x^{\frac{3}{2}}}} \right]_2^4\\ & = 2\left[ {{{\left( 4 \right)}^{\frac{3}{2}}} - {{\left( 2 \right)}^{\frac{3}{2}}}} \right]\\ &= 2\left[ {8 - 2\sqrt 2 } \right]\\ &= \left( {16 - 4\sqrt 2 } \right)\end{align}

## Chapter 8 Ex.8.1 Question 3

Find the area of the region bounded by $${x^2} = 4y,y = 2,y = 4$$ and the y-axis in the first quadrant.

### Solution

\begin{align}ar\left( {ABCD} \right) &= \int_2^4 {xdy} \\ &= \int_2^4 {2\sqrt y dy} = 2\int_2^4 {\sqrt y dy} \\& = 2\left[ {\frac{{{y^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_2^4\\ &= \frac{4}{3}\left[ {{{\left( 4 \right)}^{\frac{3}{2}}} - {{\left( 2 \right)}^{\frac{3}{2}}}} \right] = \frac{4}{3}\left[ {8 - 2\sqrt 2 } \right]\\& = \left( {\frac{{32 - 8\sqrt 2 }}{3}} \right)\end{align}

## Chapter 8 Ex.8.1 Question 4

Find the area of the region bounded by the ellipse $$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$$.

### Solution

It is given that

\begin{align}& \Rightarrow \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\\ &\Rightarrow \frac{{{y^2}}}{9} = 1 - \frac{{{x^2}}}{{16}}\\ &\Rightarrow y = 3\sqrt {1 - \frac{{{x^2}}}{{16}}}\end{align}

Area of ellipse $$= 4 \times ar\left( {OAB} \right)$$

\begin{align}ar\left( {OAB} \right)& = \int_0^4 {ydx} \\ &= \int_0^4 {3\sqrt {1 - \frac{{{x^2}}}{{16}}} dx} \\ &= \frac{3}{4}\int_0^4 {\sqrt {16 - {x^2}} } dx\\ &= \frac{3}{4}\left[ {\frac{x}{2}\sqrt {16 - {x^2}} + \frac{{16}}{2}{{\sin }^{ - 1}}\frac{x}{4}} \right]_0^4\\ &= \frac{3}{4}\left[ {2\sqrt {16 - 16} + 8{{\sin }^{ - 1}}\left( 1 \right) - 0 - 8{{\sin }^{ - 1}}\left( 0 \right)} \right]\\ &= \frac{3}{4}\left[ {\frac{{8\pi }}{2}} \right]\\ &= \frac{3}{4}\left[ {4\pi } \right]\\ &= 3\pi\end{align}

Area of ellipse $$= 4 \times 3\pi = 12\pi$$ units

## Chapter 8 Ex.8.1 Question 5

Find the area of the region bounded by the ellipse $$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$$

### Solution

It is given that

\begin{align}& \Rightarrow \frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1\\& \Rightarrow y = 3\sqrt {1 - \frac{{{x^2}}}{4}}\end{align}

Area of ellipse $$= 4 \times ar\left( {OAB} \right)$$

\begin{align}ar\left( {OAB} \right) &= \int_0^2 {ydx} \\&= \int_0^2 {3\sqrt {1 - \frac{{{x^2}}}{4}} } dx\\ &= \frac{3}{2}\int_0^2 {\sqrt {4 - {x^2}} } dx\\& = \frac{3}{2}\left[ {\frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{{\sin }^{ - 1}}\frac{x}{2}} \right]_0^2\\ &= \frac{3}{2}\left[ {\frac{{2\pi }}{2}} \right]\\ &= \frac{{3\pi }}{2}\end{align}

Area of ellipse $$= 4 \times \frac{{3\pi }}{2} = 6\pi$$ units.

## Chapter 8 Ex.8.1 Question 6

Find the area of the region in the first quadrant enclosed by x-axis, line $$x = \sqrt 3 y$$ and the circle $${x^2} + {y^2} = 4$$

### Solution

$$ar\left( {OAB} \right) = ar\left( {\Delta OAC} \right) + ar\left( {ABC} \right)$$

\begin{align}ar\left( {\Delta OAC} \right) &= \frac{1}{2} \times OC \times AC\\ &= \frac{1}{2} \times \sqrt 3 \times 1\\ &= \frac{{\sqrt 3 }}{2}\end{align}

\begin{align}ar\left( {ABC} \right)&= \int_{\sqrt 3 }^2 {ydx} \\ &= \int_{\sqrt 3 }^2 {\sqrt {4 - {x^2}} dx} \\ &= \left[ {\frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{{\sin }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_{\sqrt 3 }^2\\ &= \left[ {2 \times \frac{\pi }{2} - \frac{{\sqrt 3 }}{2}\sqrt {4 - 3} - 2{{\sin }^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)} \right]\\& = \left[ {\pi - \frac{{\sqrt 3 }}{2} - 2\left( {\frac{\pi }{3}} \right)} \right]\\ &= \left[ {\pi - \frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}} \right]\\& = \left[ {\frac{{3\pi - 2\pi }}{3} - \frac{{\sqrt 3 }}{2}} \right]\\ &= \left[ {\frac{\pi }{3} - \frac{{\sqrt 3 }}{2}} \right]\end{align}

Therefore, required area enclosed $$= \frac{{\sqrt 3 }}{2} + \frac{\pi }{3} - \frac{{\sqrt 3 }}{2} = \frac{\pi }{3}$$ square units.

## Chapter 8 Ex.8.1 Question 7

Find the area of the smaller part of the circle $${x^2} + {y^2} = {a^2}$$ cut off by the line $$x = \frac{a}{{\sqrt 2 }}$$.

### Solution

The area of the smaller part of the circle, $${x^2} + {y^2} = {a^2}$$ cut off by the line, $$x = \frac{a}{{\sqrt 2 }}$$, is the area ABCD.

It can be observed that the area ABCD is symmetrical about x-axis.

$$ar\left( {ABCD} \right) = 2 \times ar\left( {ABC} \right)$$

\begin{align}ar\left( {ABC} \right) &= \int_{\frac{a}{{\sqrt 2 }}}^a {ydx} \\ &= \int_{\frac{a}{{\sqrt 2 }}}^a {\sqrt {{a^2} - {x^2}} dx} \\ &= \left[ {\frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{a}} \right)} \right]_{\frac{a}{{\sqrt 2 }}}^a\\& = \left[ {\frac{{{a^2}}}{2}\left( {\frac{\pi }{2}} \right) - \frac{a}{{2\sqrt 2 }}\sqrt {{a^2} - \frac{{{a^2}}}{2}} - \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)} \right]\\ &= \frac{{{a^2}\pi }}{4} - \frac{a}{{2\sqrt 2 }}.\frac{a}{{\sqrt 2 }} - \frac{{{a^2}}}{2}\left( {\frac{\pi }{4}} \right)\\ &= \frac{{{a^2}\pi }}{4} - \frac{{{a^2}}}{4} - \frac{{{a^2}\pi }}{8}\\ &= \frac{{{a^2}}}{4}\left[ {\pi - 1 - \frac{\pi }{2}} \right]\\& = \frac{{{a^2}}}{4}\left[ {\frac{\pi }{2} - 1} \right]\end{align}

\begin{align}ar\left( {ABCD} \right) &= 2\left[ {\frac{{{a^2}}}{4}\left( {\frac{\pi }{2} - 1} \right)} \right]\\ &= \frac{{{a^2}}}{2}\left( {\frac{\pi }{2} - 1} \right)\end{align}

Therefore, the required area is $$\frac{{{a^2}}}{2}\left( {\frac{\pi }{2} - 1} \right)$$ square units.

## Chapter 8 Ex.8.1 Question 8

The area between $$x = {y^2}$$ and $$x = {\rm{4 }}$$ is divided into two equal parts by the line $$x = a$$, find the value of $$a$$.

### Solution

The line $$x = a$$ divides the area bounded by the parabola and $$x = {\rm{4 }}$$into two equal parts.

Therefore, $$ar\left( {OAD} \right) = ar\left( {ABCD} \right)$$

It can be observed that the given area is symmetrical about x-axis.

Hence, $$ar\left( {OED} \right) = ar\left( {EFCD} \right)$$

\begin{align}ar\left( {OED} \right) &= \int_0^a {ydx} \\ = \int_0^a {\sqrt x dx} \\ &= \left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^a\\ = \frac{2}{3}{a^{\frac{3}{2}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

\begin{align}ar\left( {EFCD} \right) &= \int_a^4 {\sqrt x dx} \\ &= \left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_a^4\\ &= \frac{2}{3}\left[ {8 - {a^{\frac{3}{2}}}} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From (1) and (2), we obtain

\begin{align}& \Rightarrow \frac{2}{3}{\left( a \right)^{\frac{3}{2}}} = \frac{2}{3}\left[ {8 - {{\left( a \right)}^{\frac{3}{2}}}} \right]\\& \Rightarrow 2{\left( a \right)^{\frac{3}{2}}} = 8\\& \Rightarrow {\left( a \right)^{\frac{3}{2}}} = 4\\& \Rightarrow a = {\left( 4 \right)^{\frac{2}{3}}}\end{align}

Therefore, the value of $$a = {\left( 4 \right)^{\frac{2}{3}}}$$.

## Chapter 8 Ex.8.1 Question 9

Find the area of the region bounded by the parabola $$y = {x^2}$$ and the line $$y = \left| x \right|$$.

### Solution

The area bounded by the parabola $$y = {x^2}$$ and the line $$y = \left| x \right|$$, can be represented as

The given area is symmetrical about y-axis.

Therefore, $$ar\left( {OACO} \right) = ar\left( {ODBO} \right)$$

The point of intersection of parabola $$y = {x^2}$$ and the line $$y = \left| x \right|$$, is $$A\left( {1,1} \right)$$.

$$ar\left( {OACO} \right) = ar\left( {\Delta OAM} \right) - ar\left( {OMACO} \right)$$

\begin{align}ar\left( {OMACO} \right) &= \int_0^1 {ydx} \\ &= \int_0^1 {{x^2}dx} \\ &= \left[ {\frac{{{x^3}}}{3}} \right]_0^1\\ &= \frac{1}{3}\end{align}

\begin{align}ar\left( {OACO} \right) &= ar\left( {\Delta OAM} \right) - ar\left( {OMACO} \right)\\& = \frac{1}{2} - \frac{1}{3}\\ &= \frac{1}{6}\end{align}

Therefore, the required area $$= 2\left[ {\frac{1}{6}} \right] = \frac{1}{3}$$ units.

## Chapter 8 Ex.8.1 Question 10

Find the area bounded by the curve $${x^2} = 4y$$ and the line $$x = 4y - 2$$.

### Solution

Coordinates of point $$A\left( { - 1,\frac{1}{4}} \right)$$.

Coordinates of point $$B\left( {2,1} \right)$$.

Draw AL and BM perpendicular to x-axis.

$$ar\left( {OBAO} \right) = ar\left( {OBCO} \right) + ar\left( {OACO} \right)$$

\begin{align}ar\left( {OBCO} \right) &= ar\left( {OMBC} \right) - ar\left( {OMBO} \right)\\ &= \int_0^2 {\frac{{x + 2}}{4}dx - \int_0^2 {\frac{{{x^2}}}{4}dx} } \\ &= \frac{1}{4}\left[ {\frac{{{x^2}}}{2} + 2x} \right]_0^2 - \frac{1}{4}\left[ {\frac{{{x^3}}}{3}} \right]_0^2\\& = \frac{1}{4}\left[ {2 + 4} \right] - \frac{1}{4}\left[ {\frac{8}{3}} \right]\\& = \frac{3}{2} - \frac{2}{3}\\& = \frac{5}{6}\end{align}

\begin{align}ar\left( {OACO} \right) &= ar\left( {OLAC} \right) - ar\left( {OLAO} \right)\\& = \int_{ - 1}^0 {\frac{{x + 2}}{4}dx} - \int_{ - 1}^0 {\frac{{{x^2}}}{4}dx} \\& = \frac{1}{4}\left[ {\frac{{{x^2}}}{2} + 2x} \right]_{ - 1}^0 - \frac{1}{4}\left[ {\frac{{{x^3}}}{3}} \right]_{ - 1}^0\\ &= - \frac{1}{4}\left[ {\frac{{{{\left( { - 1} \right)}^2}}}{2} + 2\left( { - 1} \right)} \right] - \left[ { - \frac{1}{4}\left( {\frac{{{{\left( { - 1} \right)}^3}}}{3}} \right)} \right]\\ &= - \frac{1}{4}\left[ {\frac{1}{2} - 2} \right] - \frac{1}{{12}}\\& = - \frac{1}{8} + \frac{1}{2} - \frac{1}{{12}}\\ &= \frac{7}{{24}}\end{align}

Required area $$= \left( {\frac{5}{6} + \frac{7}{{24}}} \right) = \frac{9}{8}$$ units.

## Chapter 8 Ex.8.1 Question 11

Find the area of the region bounded by the curve $${y^2} = 4x$$ and the line $$x = 3$$.

### Solution

Therefore, $$ar\left( {OACO} \right) = 2 \times ar\left( {AOB} \right)$$

\begin{align}ar\left( {OACO} \right) &= 2\left[ {\int_0^3 {ydx} } \right]\\ &= 2\left[ {\int_0^3 {2\sqrt x dx} } \right]\\& = 4\left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^3\\ &= \frac{8}{3}\left[ {{{\left( 3 \right)}^{\frac{3}{2}}}} \right]\\ &= 8\sqrt 3\end{align}

Required area is $$8\sqrt 3$$ units.

## Chapter 8 Ex.8.1 Question 12

Area lying in the first quadrant and bounded by the circle $${x^2} + {y^2} = 4$$ and the lines $$x = 0$$ and $$x = {\rm{2}}$$ is

(A) $$\pi$$

(B) $$\frac{\pi }{2}$$

(C) $$\frac{\pi }{3}$$

(D) $$\frac{\pi }{4}$$

### Solution

\begin{align}ar\left( {OAB} \right) &= \int_0^2 {ydx} \\& = \int_0^2 {\sqrt {4 - {x^2}} } dx\\ &= \left[ {\frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{{\sin }^{ - 1}}\frac{x}{2}} \right]_0^2\\& = 2\left( {\frac{\pi }{2}} \right)\\& = \pi\end{align}

## Chapter 8 Ex.8.1 Question 13

Area of the region bounded by the curve $${y^2} = 4x$$, $$y$$-axis and the line $$y = 3$$ is

(A) $$2$$

(B) $$\frac{9}{4}$$

(C) $$\frac{9}{3}$$

(D) $$\frac{9}{2}$$

### Solution

\begin{align}ar\left( {OAB} \right) &= \int_0^3 {xdy} \\& = \int_0^3 {\frac{{{y^2}}}{4}dy} \\& = \frac{1}{4}\left[ {\frac{{{y^3}}}{3}} \right]_0^3\\ &= \frac{1}{{12}}\left( {27} \right)\\& = \frac{9}{4}\end{align}

Correct answer is $$B$$.

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