Science:Math Exam Resources/Courses/MATH200/April 2012/Question 09

MATH200 April 2012

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Question 09
Let E be the smaller of the two solid regions bounded by the surfaces $\displaystyle {}z=x^{2}+y^{2}$ and $\displaystyle {}x^{2}+y^{2}+z^{2}=6.$ Evaluate $\iiint \limits _{E}(x^{2}+y^{2}){\textrm {d}}V.$

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
Make a sketch of the sphere and the paraboloid to find the region of integration.

Hint 2
According to the figure in Hint 1, the region of integration is bounded from below by the paraboloid $\displaystyle z=x^{2}+y^{2}$ and from above by the sphere $\displaystyle x^{2}+y^{2}+z^{2}=6$

Hint 3
Switch to cylindrical coordinates and calculate the intersection of the paraboloid and the sphere to find the upper integration boundary 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 region of integration is bounded from above by $\displaystyle z=x^{2}+y^{2}$ and from below by $\displaystyle x^{2}+y^{2}+z^{2}=6$. The intersection of the paraboloid and the sphere is found by substituting the first equation into the second, and solving for z: $\displaystyle x^{2}+y^{2}+z^{2}=z+z^{2}=6$, hence $\displaystyle z_{1,2}=2,-3$. Since we see on the sketch from Hint 1, that the two surfaces intersect at a positive $\displaystyle z$ value, the intersection is at $\displaystyle z=2$. We switch to cylindrical coordinates and express the $\displaystyle z$-variable in dependence of the radius $\displaystyle r={\sqrt {x^{2}+y^{2}}}$. The largest value of $\displaystyle r$ in the region of integration is where the sphere and the paraboloid intersect, namely at $\displaystyle r={\sqrt {z}}={\sqrt {2}}$. The integration limits for the radius are $\displaystyle 0\leq r\leq {\sqrt {2}}$. The variable $\displaystyle z$ is bounded from below by $\displaystyle z=r^{2}$ and from above by $\displaystyle r^{2}+z^{2}=6$, that is, $\displaystyle z={\sqrt {6-r^{2}}}$. So the integration boundaries are $\displaystyle r^{2}\leq z\leq {\sqrt {6-r^{2}}}$. The angle $\displaystyle \phi$ does not depend on $\displaystyle z$ or $\displaystyle r$ and completes the whole circle $\displaystyle 0\leq \phi \leq 2\pi$. The integrand function in cylindrical coordinates is $\displaystyle x^{2}+y^{2}=r^{2}$ Hence, the integral to calculate is $\displaystyle {\begin{aligned}\int _{0}^{\sqrt {2}}\int _{r^{2}}^{\sqrt {6-r^{2}}}\int _{0}^{2\pi }r^{2}r\;{\text{d}}\phi {\text{d}}z{\text{d}}r=&\int _{0}^{\sqrt {2}}\int _{r^{2}}^{\sqrt {6-r^{2}}}2\pi r^{3}\;{\text{d}}z{\text{d}}r\\=&\int _{0}^{\sqrt {2}}2\pi r^{3}({\sqrt {6-r^{2}}}-r^{2})\;{\text{d}}r\\=&\int _{0}^{\sqrt {2}}2\pi r^{3}(6-r^{2})^{\frac {1}{2}}\;{\text{d}}r-\int _{0}^{\sqrt {2}}2\pi r^{5}\;{\text{d}}r\end{aligned}}$ Now we perform integration by parts in the first integral, where we consider $\displaystyle u=2\pi r^{2}$ and $\displaystyle dv=r(6-r^{2})^{\frac {1}{2}}$. Then $\displaystyle du=4\pi r$ and $\displaystyle v=-{\frac {1}{3}}(6-r^{2})^{\frac {3}{2}}$. Overall we obtain $\displaystyle {\begin{aligned}=&\left[-2\pi r^{2}{\frac {1}{3}}(6-r^{2})^{\frac {3}{2}}\right]_{0}^{\sqrt {2}}+\int _{0}^{\sqrt {2}}4\pi r{\frac {1}{3}}(6-r^{2})^{\frac {3}{2}}\;{\text{d}}r-\left[2\pi {\frac {r^{6}}{6}}\right]_{0}^{\sqrt {2}}\\=&-2\pi 2{\frac {1}{3}}(6-2)^{\frac {3}{2}}+0-\left[4\pi {\frac {1}{15}}(6-r^{2})^{\frac {5}{2}}\right]_{0}^{\sqrt {2}}-2\pi {\sqrt {2}}^{6}{\frac {1}{6}}+0\\=&-{\frac {32}{3}}\pi -{\frac {4}{15}}\pi 32+{\frac {4}{15}}\pi 36{\sqrt {6}}-{\frac {8}{3}}\pi \\=&{\frac {\pi }{15}}(144{\sqrt {6}}-328)\approx 5.18\end{aligned}}$

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