Convert the integral to polar coordinates. $\int _ { - 1 } ^ { 1 } \int _ { 1 } ^ { \sqrt { 2 - x ^ { 2 } } } f ( x , y ) d y d x$
A) $\int _ { \pi / 4 } ^ { 3 \pi / 4 } \int _ { 1 / \sin \theta } ^ { \sqrt { 2 } } f ( r \cos \theta , r \sin \theta ) d r d \theta$
B) $\int _ { 0 } ^ { 3 \pi / 4 } \int _ { 1 / \sin \theta } ^ { \sqrt { 2 } } f ( r \cos \theta , r \sin \theta ) r d r d \theta$
C) $\int _ { \pi / 4 } ^ { 3 \pi / 4 } \int _ { 1 } ^ { \sqrt { 2 } } f ( r \cos \theta , r \sin \theta ) r d r d \theta$
D) $\int _ { \pi / 4 } ^ { 3 \pi / 4 } \int _ { 1 / \sin \theta } ^ { \sqrt { 2 } } f ( r \cos \theta , r \sin \theta ) r d r d \theta$
E) $\int _ { \pi / 4 } ^ { 3 \pi / 4 } \int _ { 1 / \sin \theta } ^ { \sqrt { 2 } } f ( r \cos \theta , r \sin \theta ) r d \theta d r$

Question

Quizplus · Accepted Answer

The correct conversion to polar coordinates involves setting the limits for $\theta$ based on the symmetry of the region, which is from $\pi/4$ to $3\pi/4$ to cover the upper half of the unit circle to the circle of radius $\sqrt{2}$. The radial limits are from $1/\sin\theta$ (to ensure $y \geq 1$) to $\sqrt{2}$, and the inclusion of $r$ in the differential area element $r dr d\theta$ is necessary for polar integration.

Convert the Integral to Polar Coordinates ∫−11∫12−x2f(x,y)dydx\int _ { - 1 } ^ { 1 } \int _ { 1 } ^ { \sqrt { 2 - x ^ { 2 } } } f ( x , y ) d y d x∫−11​∫12−x2​​f(x,y)dydx

Convert the Integral to Polar Coordinates $\int _ { - 1 } ^ { 1 } \int _ { 1 } ^ { \sqrt { 2 - x ^ { 2 } } } f ( x , y ) d y d x$