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

MATH200 December 2011

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  •  April 2012  •  December 2011  •  

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 $\displaystyle 4\pi r^{3}/3$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

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:

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 $\displaystyle x^{2}+y^{2}+z^{2}=12$.

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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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\displaystyle V_{b}+V_{h}-V_{i}=\pi \left(36+16{\sqrt {3}}\right)$

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