If R is the region bounded by $x ^ { 2 } + y ^ { 2 } = 9 , x \geq 0 , y \geq 0$ then $\iint$_R $( 2 x + 5 y ) d A$ in polar form is A) $\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } ( 2 \cos \theta + 5 \sin \theta ) d r \right] d \theta$ B) $\int _ { - \pi } ^ { \pi } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } ( 2 \cos \theta + 5 \sin \theta ) d r \right] d \theta$ C) $\int _ { 0 } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } ( 2 \cos \theta + 5 \sin \theta ) d r \right] d \theta$ D) $\int _ { 0 } ^ { \pi } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } ( 2 \cos \theta + 5 \sin \theta ) d r \right] d \theta$ E) $\int _ { 0 } ^ { 2 \pi } \left[ \int _ { 0 } ^ { 3 } r ^ { 2 } ( 2 \cos \theta + 5 \sin \theta ) d r \right] d \theta$

Question

Quizplus · Accepted Answer

The region R is a quarter circle in the first quadrant, so the limits for $\theta$ are from 0 to $\frac{\pi}{2}$. The radius r goes from 0 to 3, the radius of the circle. The integrand is correctly converted to polar coordinates, where $x = r\cos\theta$ and $y = r\sin\theta$.

If R Is the Region Bounded By x2+y2=9,x≥0,y≥0x ^ { 2 } + y ^ { 2 } = 9 , x \geq 0 , y \geq 0x2+y2=9,x≥0,y≥0

If R Is the Region Bounded By $x ^ { 2 } + y ^ { 2 } = 9 , x \geq 0 , y \geq 0$