Class 12 Maths NCERT Solutions Chapter 8 Ex 8.1 – Area Under Simple Curves

Here a^2=16 \Rightarrow a=4 and b^2=9 \Rightarrow b=3. Since the ellipse is symmetrical about both axes, the required area is 4 times the area in the first quadrant.

Solving for y: y=\dfrac{3}{4}\sqrt{16-x^2}, taking the positive value as the region lies in the first quadrant.

Taking vertical strips: \text{Area}=4\displaystyle\int_{0}^{4} y\,dx=4\displaystyle\int_{0}^{4}\dfrac{3}{4}\sqrt{16-x^2}\,dx=3\displaystyle\int_{0}^{4}\sqrt{16-x^2}\,dx.

Using \displaystyle\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} with a=4:

\text{Area}=3\left[\dfrac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\dfrac{x}{4}\right]_0^4=3\left[\left(0+8\cdot\dfrac{\pi}{2}\right)-0\right]=3(4\pi).

Here the ellipse meets the x-axis at x=\pm 2 and the y-axis at y=\pm 3. Since it's symmetrical about both axes, the required area is 4 times the area in the first quadrant, taking vertical strips of width dx from x=0 to x=2.

Solving for y: y=\dfrac{3}{2}\sqrt{4-x^2}, taking the positive value as the region lies in the first quadrant.

\text{Area}=4\displaystyle\int_{0}^{2} y\,dx=4\displaystyle\int_{0}^{2}\dfrac{3}{2}\sqrt{4-x^2}\,dx=6\displaystyle\int_{0}^{2}\sqrt{4-x^2}\,dx.

Using \displaystyle\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} with a=2:

\text{Area}=6\left[\dfrac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\dfrac{x}{2}\right]_0^2=6\left[\left(0+2\cdot\dfrac{\pi}{2}\right)-0\right]=6\pi.

The circle x^2+y^2=4 has radius 2, so it passes through (2,0) — the line x=2 is a tangent-ordinate right at the edge of the circle. So the required region is exactly the part of the circle in the first quadrant, between x=0 and x=2.

Solving for y in the first quadrant: y=\sqrt{4-x^2}.

\text{Area}=\displaystyle\int_{0}^{2}\sqrt{4-x^2}\,dx=\left[\dfrac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\dfrac{x}{2}\right]_0^2.

=\left(0+2\cdot\dfrac{\pi}{2}\right)-\left(0+2\sin^{-1}0\right)=\pi-0=\pi.

Since the boundary is given in terms of y (the y-axis and a horizontal line y=3), it's natural to take horizontal strips and integrate with respect to y.

From y^2=4x: x=\dfrac{y^2}{4}.

\text{Area}=\displaystyle\int_{0}^{3} x\,dy=\displaystyle\int_{0}^{3}\dfrac{y^2}{4}\,dy=\dfrac{1}{4}\left[\dfrac{y^3}{3}\right]_0^3=\dfrac{1}{4}\left(\dfrac{27}{3}\right)=\dfrac{1}{4}(9)=\dfrac{9}{4}.