The area of the region bounded by frac{x^2}{9 | vedclass

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(B) The given equations are the ellipse $\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$. Let $x = 3 \cos 	heta$ and $y =

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

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(B) The given equations are the ellipse $\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$. Let $x = 3 \cos 	heta$ and $y = 2 \sin 	heta$. The line equation becomes $\cos 	heta + \sin 	heta = 1$. Solving for $	heta$,we get $	heta = 0$ or $	heta = \frac{\pi}{2}$. The area of the region bounded by the ellipse and the line is the area of the sector of the ellipse minus the area of the triangle formed by the line and the axes. The area of the ellipse is $\pi ab = \pi(3)(2) = 6\pi$. The area of the region in the first quadrant bounded by the ellipse is $\frac{1}{4}(6\pi) = \frac{3\pi}{2}$. The area of the triangle formed by the line $\frac{x}{3}+\frac{y}{2}=1$ with the axes is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 3 	imes 2 = 3$. Thus,the required area is $\frac{3\pi}{2} - 3 = \frac{3}{2}(\pi - 2)$ sq. units.

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

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