Area of the region enclosed between the curve | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The curves are $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$.Since both curves are symmetric about the $x$-axis,the total area $A$ is twice the area

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The curves are $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$.Since both curves are symmetric about the $x$-axis,the total area $A$ is twice the area in the first quadrant where $y \in [0, 1]$.In the first quadrant,$x = y \sqrt{1 - y^2}$ and $x = y^2 - 1$.The area $A$ is given by:$A = 2 \int_{0}^{1} [y \sqrt{1 - y^2} - (y^2 - 1)] \, dy$$A = 2 \left[ \int_{0}^{1} y \sqrt{1 - y^2} \, dy - \int_{0}^{1} (y^2 - 1) \, dy \right]$Let $u = 1 - y^2$,then $du = -2y \, dy$,so $y \, dy = -\frac{1}{2} du$.When $y=0, u=1$; when $y=1, u=0$.$\int_{0}^{1} y \sqrt{1 - y^2} \, dy = -\frac{1}{2} \int_{1}^{0} \sqrt{u} \, du = \frac{1}{2} \int_{0}^{1} u^{1/2} \, du = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{1} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$.$\int_{0}^{1} (y^2 - 1) \, dy = \left[ \frac{y^3}{3} - y \right]_{0}^{1} = \frac{1}{3} - 1 = -\frac{2}{3}$.Therefore,$A = 2 \left[ \frac{1}{3} - (-\frac{2}{3}) \right] = 2 \left[ \frac{1}{3} + \frac{2}{3} \right] = 2(1) = 2$.

Area of the region enclosed between the curves $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$ is

Similar Questions

The area of the region between the curves $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac{\pi}{4}$ is

The area bounded by the curves $y = \sqrt{x}$,$2y + 3 = x$ and the $x$-axis in the $1^{st}$ quadrant is

The area bounded by the curves $y = |x| - 1$ and $y = -|x| + 1$ is

The area of the region enclosed by the curves $y=x^2-4x+4$ and $y^2=16-8x$ is:

The area (in sq. units) of the region $A = \{(x, y) : x^2 \le y \le x + 2\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module