Solve the problem.
-Let $D$ be the region bounded below by the $x y$-plane, above by the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 49$, and on the sides by the cylinder $x ^ { 2 } + y ^ { 2 } = 36$. Set up the triple integral in cylindrical coordinates that gives the volume of $D$ using the order of integration $\mathrm { dz } \mathrm { dr } \mathrm { d } \theta$.
A) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 7 } \int _ { 0 } ^ { \sqrt { 36 - z ^ { 2 } } } r d z d r d \theta$
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 6 } \int _ { 0 } ^ { \sqrt { 49 - z ^ { 2 } } } r d z d r d \theta$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 7 } \int _ { 0 } ^ { \sqrt { 36 - r ^ { 2 } } } r d z d r d \theta$
D) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 6 } \int _ { 0 } ^ { \sqrt { 49 - r ^ { 2 } } } r d z d r d \theta$

Question

Quizplus · Accepted Answer

The volume of $D$ is found by integrating over the cylindrical coordinates. The limits for $\theta$ are from $0$ to $2\pi$ since the region is symmetric around the z-axis. The radius $r$ goes from $0$ to $6$ because the cylinder's equation is $x^2 + y^2 = 36$. The height $z$ varies from $0$ (the xy-plane) to the top of the sphere, which is given by the equation $z = \sqrt{49 - r^2}$ after rearranging the sphere's equation $x^2 + y^2 + z^2 = 49$ for $z$. Therefore, the correct integral setup is $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 6 } \int _ { 0 } ^ { \sqrt { 49 - r ^ { 2 } } } r d z d r d \theta$.

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