Question

The area bounded by the ellipse $x^2+9y^2=9$ and the straight line $x+3y=3$ is

• A. $3\pi$
• B. $4\pi$
• C. $6\pi$
• D. $9\pi$

Answer 1

The correct option is A $3\pi$

The intersection points of $x^2+9x^2=9$ and $x+3y=3$ are $(0,1)$ and $(3,0)$.

Let $A={\int }_0^1\pi x^{2}dy-{\int }_0^1\pi x^{2}dy$

$\Rightarrow A={\int }_0^1\pi (9-9y^2)dy-{\int }_0^1\pi 9{(1-y)}^{2}dy$

$\Rightarrow A=9\pi {[y-\frac{y^3}{3}]}_0^1-9\pi {\left[\frac{{(1-y)}^3}{-3}\right]}_0^1$

$\Rightarrow A=9\pi (1-\frac{1}{3})-\frac{9\pi }{3}(0-1)$

$\therefore A=6\pi -3\pi =3\pi .$