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

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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.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

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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