Solve the problem.
-Let D be the smaller cap cut from a solid ball of radius 9 units by a plane 8 units from the center of the sphere. Set up the triple integral for the volume of D in cylindrical coordinates.
A) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 145 } } \int _ { 9 } ^ { \sqrt { 64 - r ^ { 2 } } } r d z d r d \theta$
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 17 } } \int _ { 9 } ^ { \sqrt { 64 - r ^ { 2 } } } \mathrm { rdzdrd \theta }$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 17 } } \int _ { 8 } ^ { \sqrt { 81 - r ^ { 2 } } } r d z d r d \theta$
D) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 145 } } \int _ { 8 } ^ { \sqrt { 81 - r ^ { 2 } } } r d z d r d \theta$

Question

Solve the problem.
-Let D be the smaller cap cut from a solid ball of radius 9 units by a plane 8 units from the center of the sphere. Set up the triple integral for the volume of D in cylindrical coordinates. 
A) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 145 } } \int _ { 9 } ^ { \sqrt { 64 - r ^ { 2 } } } r d z d r d \theta$
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 17 } } \int _ { 9 } ^ { \sqrt { 64 - r ^ { 2 } } } \mathrm { rdzdrd \theta }$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 17 } } \int _ { 8 } ^ { \sqrt { 81 - r ^ { 2 } } } r d z d r d \theta$
D) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 145 } } \int _ { 8 } ^ { \sqrt { 81 - r ^ { 2 } } } r d z d r d \theta$

Quizplus · Accepted Answer

The volume of the cap D can be found using cylindrical coordinates. The limits for $\theta$ are from 0 to $2\pi$ since it's a full rotation around the z-axis. The radius $r$ of the cap at the base is determined by the intersection of the plane and the sphere, which forms a circle. Using the Pythagorean theorem in the context of the sphere's equation $x^2 + y^2 + z^2 = 81$ (since the radius of the sphere is 9), and knowing the plane cuts the sphere 8 units from the center, we find the radius of this circle by setting $z = 8$ and solving for $r$, which gives $r = \sqrt{81 - 8^2} = \sqrt{17}$. Thus, the limits for $r$ are from 0 to $\sqrt{17}$. The limits for $z$ are from the plane cutting the sphere (z = 8) up to the surface of the sphere, which is described by $z = \sqrt{81 - r^2}$. Therefore, the correct integral setup is $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 17 } } \int _ { 8 } ^ { \sqrt { 81 - r ^ { 2 } } } r d z d r d \theta$.

Solve the Problem ∫02π∫0145∫964−r2rdzdrdθ\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 145 } } \int _ { 9 } ^ { \sqrt { 64 - r ^ { 2 } } } r d z d r d \theta∫02π​∫0145​​∫964−r2​​rdzdrdθ

Solve the Problem $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 145 } } \int _ { 9 } ^ { \sqrt { 64 - r ^ { 2 } } } r d z d r d \theta$