Find the area of the smaller region bounded b | vedclass

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

Question

(A) The area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is the area of

Vedclass · Accepted Answer

$\frac{3}{2}(\pi-2)$ sq. units

Vedclass · Answer

(A) The area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is the area of the region $ABC$ shown in the figure.The points of intersection are $A(0, 2)$ and $B(3, 0)$.Area of the region $ABC = \int_{0}^{3} (y_{	ext{ellipse}} - y_{	ext{line}}) dx$From the ellipse equation,$y = 2\sqrt{1-\frac{x^{2}}{9}} = \frac{2}{3}\sqrt{9-x^{2}}$.From the line equation,$y = 2(1-\frac{x}{3}) = 2 - \frac{2x}{3}$.Area $= \int_{0}^{3} \left( \frac{2}{3}\sqrt{9-x^{2}} - (2 - \frac{2x}{3}) ight) dx$$= \frac{2}{3} \int_{0}^{3} \sqrt{3^{2}-x^{2}} dx - \int_{0}^{3} (2 - \frac{2x}{3}) dx$$= \frac{2}{3} \left[ \frac{x}{2}\sqrt{9-x^{2}} + \frac{9}{2}\sin^{-1}(\frac{x}{3}) ight]_{0}^{3} - \left[ 2x - \frac{x^{2}}{3} ight]_{0}^{3}$$= \frac{2}{3} \left[ (0 + \frac{9}{2} \cdot \frac{\pi}{2}) - 0 ight] - \left[ (6 - 3) - 0 ight]$$= \frac{2}{3} \cdot \frac{9\pi}{4} - 3 = \frac{3\pi}{2} - 3 = \frac{3}{2}(\pi-2)$ sq. units.

Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$.

Similar Questions

The area in square units of the region bounded by the curve $y = \sqrt{16 - x^2}$ and the lines $x = 0, x = 4$ above the $X$-axis is: (in $\pi$)

The line $y = mx$ bisects the area enclosed by the curve $y = 1 + 4x - x^2$ and the lines $x = 0, x = \frac{3}{2}$ and $y = 0$. Then the value of $m$ is:

The area bounded by the circle $x^{2}+y^{2}=16$ and the lines $x=0$ and $x=2$ is

The area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $x = 1$ is...

The area between the curve $y = 4 + 3x - x^2$ and the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module