$$\text { Set up the iterated integral for evaluating } \iint _ { D } \int f ( r , \theta , z ) d z r d r d \theta \text { over the given region D. }$$
-$\mathrm { D }$ is the solid right cylinder whose base is the region in the xy-plane that lies inside the cardioid $r = 6 - 2$ sin $\theta$ and outside the circle $r = 4$, and whose top lies in the plane $z = 10$.
A) $\int_{0}^{\pi} \int_{4}^{6-2 \sin \theta} \int_{0}^{10} \mathrm{f}(\mathrm{r}, \theta, \mathrm{z}) \mathrm{dz} \mathrm{rdr} \mathrm{d} \theta$
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 4 } ^ { 6 - 2 \sin \theta } \int _ { 0 } ^ { 10 } \mathrm { f } ( \mathrm { r } , \theta , z ) \mathrm { dz } r \mathrm { dr } \mathrm { d } \theta$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 6 - 2 \sin \theta } \int _ { 0 } ^ { 10 } f ( r , \theta , z ) d z r d r d \theta$
D) $\int _ { 0 } ^ { \pi } \int _ { 0 } ^ { 6 - 2 \sin \theta } \int _ { 0 } ^ { 10 } \mathrm { f } ( \mathrm { r } , \theta , \mathrm { z } ) \mathrm { dz } r \mathrm { dr } \mathrm { d } \theta$

Question

$$\text { Set up the iterated integral for evaluating } \iint _ { D } \int f ( r , \theta , z ) d z r d r d \theta \text { over the given region D. }$$
-$\mathrm { D }$ is the solid right cylinder whose base is the region in the xy-plane that lies inside the cardioid $r = 6 - 2$ sin $\theta$ and outside the circle $r = 4$, and whose top lies in the plane $z = 10$.
A) $\int_{0}^{\pi} \int_{4}^{6-2 \sin \theta} \int_{0}^{10} \mathrm{f}(\mathrm{r}, \theta, \mathrm{z}) \mathrm{dz} \mathrm{rdr} \mathrm{d} \theta$ 
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 4 } ^ { 6 - 2 \sin \theta } \int _ { 0 } ^ { 10 } \mathrm { f } ( \mathrm { r } , \theta , z ) \mathrm { dz } r \mathrm { dr } \mathrm { d } \theta$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 6 - 2 \sin \theta } \int _ { 0 } ^ { 10 } f ( r , \theta , z ) d z r d r d \theta$
D) $\int _ { 0 } ^ { \pi } \int _ { 0 } ^ { 6 - 2 \sin \theta } \int _ { 0 } ^ { 10 } \mathrm { f } ( \mathrm { r } , \theta , \mathrm { z } ) \mathrm { dz } r \mathrm { dr } \mathrm { d } \theta$

Quizplus · Accepted Answer

The region D is a solid right cylinder whose base is a region in the xy-plane. Hence, we use cylindrical coordinates to set up the iterated integral. The given region is defined by $4 \leq r \leq 6-2\sin\theta$ and $0 \leq \theta \leq 2\pi$ and $0 \leq z \leq 10$. Therefore, the iterated integral becomes $\int _ { 0 } ^ { 2 \pi } \int _ { 4 } ^ { 6 - 2 \sin \theta } \int _ { 0 } ^ { 10 } \mathrm { f } ( \mathrm { r } , \theta , z ) \mathrm { dz } r \mathrm { dr } \mathrm { d } \theta$.