If R is the region bounded by $y = \sqrt { 9 - x ^ { 2 } }$ in the first quadrant, then $\iint$_{R $y d A$} in polar form is A) $\int _ { 0 } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 3 } r \sin \theta d r \right] d \theta$ B) $\int _ { 0 } ^ { \frac { \pi } { 4 } } \left[ \int _ { 0 } ^ { 3 } r \sin \theta d r \right] d \theta$ C) $\int _ { - \frac { \pi } { 4 } } ^ { \frac { \pi } { 4 } } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } \sin \theta d r \right] d \theta$ D) $\int _ { 0 } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } \sin \theta d r \right] d \theta$ E) $\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } \sin \theta d r \right] d \theta$

Question

Quizplus · Accepted Answer

The region R is a quarter-circle of radius 3 in the first quadrant. In polar coordinates, $y = r \sin \theta$ and the area element $dA = r dr d\theta$. The correct integral is $\int _ { 0 } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } \sin \theta d r \right] d \theta$.

If R Is the Region Bounded By y=9−x2y = \sqrt { 9 - x ^ { 2 } }y=9−x2​

If R Is the Region Bounded By $y = \sqrt { 9 - x ^ { 2 } }$