Science:Math Exam Resources/Courses/MATH200/December 2011/Question 08 (a)

MATH200 December 2011

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• April 2012 • December 2011 •

[hide] Question 08 (a)
The body of a snowman is formed by the snowballs $x^{2}+y^{2}+z^{2}=12$ (this is its body) and $x^{2}+y^{2}+(z-4)^{2}=4$ (this is its head). a) Find the volume of the snowman by subtracting the intersection of the two snow balls from the sum of the volumes of the snow balls. Recall that the volume of a sphere of radius $\displaystyle r$ is $\displaystyle 4\pi r^{3}/3$ .

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[show] Hint 1
The volume of the snowman is the volume of the body $\displaystyle V_{b}$ , the volume of the head $\displaystyle V_{h}$ minus the volume of the intesection $\displaystyle V_{i}$ . The hardest part is to find $\displaystyle V_{i}$ . To do so, we need to solve a triple integral. To find the boundaries for this triple integral, we make a sketch of the situation:

[show] Hint 2
To find $\displaystyle V_{i}$ , we use the sketch from the previous hint and see that the region of integration is bounded below from $\displaystyle x^{2}+y^{2}+(z-4)^{2}=4$ and above from $\displaystyle x^{2}+y^{2}+z^{2}=12$ .

[show] Hint 3
Finally, use cylindrical coordinates to integrate $\displaystyle \iiint _{V_{i}}1\;{\text{d}}V$ Find the intersection of the two surfaces and solve for the radius, which will give you the upper limit for the radius.

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[show] Solution
The volume of the snowman is the volume of the body $\displaystyle V_{b}$ , the volume of the head $\displaystyle V_{h}$ minus the volume of the intesection $\displaystyle V_{i}$ . First we calculate $\displaystyle V_{i}$ , which is the volume of the intersection of the two balls. The intersection is bounded below from the surface $\displaystyle x^{2}+y^{2}+(z-4)^{2}=4$ and above from $\displaystyle x^{2}+y^{2}+z^{2}=12$ . Since this shape is radially symmetric to the $\displaystyle z$ -axes, we change to cylindrical coordinates. This means $\displaystyle x^{2}+y^{2}=r^{2}$ and $\displaystyle {\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}r\cos(\phi )\\r\sin(\phi )\\z\end{pmatrix}}$ . We find the intersection of the surfaces $\displaystyle r^{2}+(z-4)^{2}=4$ and $\displaystyle r^{2}+z^{2}=12$ to find the upper bound for the radius $\displaystyle r$ : $\displaystyle 4=r^{2}+(z-4)^{2}=r^{2}+z^{2}-8z+16=12-8z+16$ which yields $\displaystyle z=3$ and $\displaystyle r={\sqrt {3}}$ . With these bounds we write $\displaystyle V_{i}$ as $\displaystyle V_{i}=\{(r,\phi ,z)\|\ 4-{\sqrt {4-r^{2}}}\leq z\leq {\sqrt {12-r^{2}}},0\leq r\leq {\sqrt {3}},0\leq \phi \leq 2\pi \}$ . Recall that the volume factor for cylindrical coordinates is $\displaystyle r$ . $\displaystyle {\begin{aligned}\iiint _{V_{i}}1\;{\text{d}}V&=\int _{0}^{\sqrt {3}}\int _{4-{\sqrt {4-r^{2}}}}^{\sqrt {12-r^{2}}}\int _{0}^{2\pi }r{\text{d}}\phi {\text{d}}z{\text{d}}r\\&=\int _{0}^{\sqrt {3}}2\pi r({\sqrt {12-r^{2}}}-4+{\sqrt {4-r^{2}}})\;{\text{d}}r\\&=\pi \left[-{\frac {2}{3}}(12-r^{2})^{\frac {3}{2}}-4r^{2}-{\frac {2}{3}}(4-r^{2})^{\frac {3}{2}}\right]_{0}^{\sqrt {3}}\\&=\pi \left[-{\frac {2}{3}}9^{\frac {3}{2}}-12-{\frac {2}{3}}+{\frac {2}{3}}12^{\frac {3}{2}}+0+{\frac {2}{3}}4^{\frac {3}{2}}\right]=\pi {\frac {4}{3}}(12{\sqrt {3}}-19)\end{aligned}}$ The ball $\displaystyle V_{b}$ which makes the body has volume $\displaystyle V_{b}={\frac {4}{3}}\pi {\sqrt {12}}^{3}=32{\sqrt {3}}\pi$ The ball $\displaystyle V_{h}$ which makes the head has volume $\displaystyle V_{h}={\frac {4}{3}}\pi 2^{3}={\frac {32}{3}}\pi$ Hence, the volume of the snowman is $\displaystyle V_{b}+V_{h}-V_{i}=\pi \left(36+16{\sqrt {3}}\right)$