The volume of a solid obtained by revolving a | vedclass

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(A) The region is bounded by the ellipse $x^2 + 9y^2 = 9$ and the line $x + 3y = 3$ in the first quadrant. For the ellipse,$x^2 = 9(1 - y^2)$. For the

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$3$

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(A) The region is bounded by the ellipse $x^2 + 9y^2 = 9$ and the line $x + 3y = 3$ in the first quadrant. For the ellipse,$x^2 = 9(1 - y^2)$. For the line,$x = 3 - 3y$,so $x^2 = (3 - 3y)^2 = 9(1 - y)^2$. The limits of integration for $y$ are from $0$ to $1$. The volume $V$ generated by revolving about the $y$-axis is given by: $V = \int_0^1 \pi x_{	ext{ellipse}}^2 dy - \int_0^1 \pi x_{	ext{line}}^2 dy$ $V = \int_0^1 \pi [9(1 - y^2)] dy - \int_0^1 \pi [9(1 - y)^2] dy$ $V = 9\pi \int_0^1 (1 - y^2 - (1 - 2y + y^2)) dy$ $V = 9\pi \int_0^1 (2y - 2y^2) dy$ $V = 18\pi \int_0^1 (y - y^2) dy$ $V = 18\pi \left[ \frac{y^2}{2} - \frac{y^3}{3} ight]_0^1$ $V = 18\pi \left( \frac{1}{2} - \frac{1}{3} ight) = 18\pi \left( \frac{1}{6} ight) = 3\pi$.

The volume of a solid obtained by revolving about the $y$-axis the region enclosed between the ellipse $x^2 + 9y^2 = 9$ and the straight line $x + 3y = 3$ in the first quadrant is: (in $\pi$)

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