Solve the problem.
-Integrate $f ( x , y ) = \sin \sqrt { x ^ { 2 } + y ^ { 2 } }$ over the region $0 \leq x ^ { 2 } + y ^ { 2 } \leq 16$.
A) $\pi ( \sin 4 - \cos 4 )$
B) $\pi ( \sin 4 - 4 \cos 4 )$
C) $2 \pi ( \sin 4 - \cos 4 )$
D) $2 \pi ( \sin 4 - 4 \cos 4 )$

Question

Quizplus · Accepted Answer

The integral is evaluated using polar coordinates, where $ r $ ranges from 0 to 4 and $\theta$ from 0 to $2\pi$. The integral becomes $\int_0^{2\pi} \int_0^4 \sin r \cdot r \, dr \, d\theta$, which evaluates to $2\pi(\sin 4 - 4\cos 4)$.

Solve the Problem f(x,y)=sin⁡x2+y2f ( x , y ) = \sin \sqrt { x ^ { 2 } + y ^ { 2 } }f(x,y)=sinx2+y2​

Solve the Problem $f ( x , y ) = \sin \sqrt { x ^ { 2 } + y ^ { 2 } }$