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

MATH200 December 2011

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Question 08 (b) iii

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). b) We can also calculate the volume of the snowman as a sum of the following triple integrals: $(1)\quad \int _{0}^{\frac {2\pi }{3}}\int _{0}^{2\pi }\int _{0}^{2}\rho ^{2}\sin(\phi )\,d\rho \,d\theta \,d\phi$ $(2)\quad \int _{0}^{2\pi }\int _{0}^{\sqrt {3}}\int _{{\sqrt {3}}r}^{4-{\frac {r}{\sqrt {3}}}}r\,dz\,dr\,d\theta$ $(3)\quad \int _{\frac {\pi }{6}}^{\pi }\int _{0}^{2\pi }\int _{0}^{2{\sqrt {3}}}\rho ^{2}\sin(\phi )\,d\rho \,d\theta \,d\phi$ Circle the right answer from the underlined choices and fill in the blanks in the following descriptions of the region of integration for each integral. [Note: We have translated the axes in order to write down some of the integrals above. The equations you specify should be those before the translation is performed.] iii. The region of integration in (3) is a part of the snowman's body / head / body and head. It is enclosed by the sphere / cone defined by the equation                       and the sphere / cone defined by the equation

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
Consider the upper and lower limits of the variables of integration. According to the size and shape of the region of integration, in particular $\displaystyle \rho \in [0,2{\sqrt {3}}]=[0,{\sqrt {12}}]$, you can identify it with the head or body of the snowman.

Hint 2

Once you found that the region of integration (3) is part of the body, the sphere enclosing the region must be the sphere equation of the body. To find the cone-equation of the cone, which is cut out from the sphere, the angle $\displaystyle \phi ={\frac {1}{6}}\pi$ is important. Rewrite the expression $\displaystyle \phi ={\frac {1}{6}}\pi$ from spherical coordinates into the equivalent expression in cartesian coordinates.

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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. Region of integration is part of head or body? In the integral (3), spherical coordinates are used, where the radius covers the interval $\displaystyle \rho \in [0,2{\sqrt {3}}]=[0,{\sqrt {12}}]$, the horizontal angle covers $\displaystyle \theta \in [0,2\pi ]$ which is the whole circle and the vertical angle covers $\displaystyle \phi \in \left[{\frac {\pi }{6}},\pi \right]$. Hence, the region of integration is a ball with radius $\displaystyle {\sqrt {12}}$, where a cone is cut out from above. Since the body of the snowman is a ball with radius $\displaystyle {\sqrt {12}}$,this region must be part of the body . Finding the equation of the enclosing sphere Since the region of integration is part of the ball with radius $\displaystyle {\sqrt {12}}$, whose centre is in the origin, it is enclosed by the sphere $\displaystyle x^{2}+y^{2}+z^{2}=12$ Finding the equation of the enclosing cone It is left to find the equation of the cone which is cut out from the region of integration. The cone is defined through the angle $\displaystyle \phi$. To find the equation of the cone, we must rewrite the expression $\displaystyle \phi ={\frac {1}{6}}\pi$ into cartesian coordinates. The relationship between spherical and cartesian coordinates is $\displaystyle {\begin{aligned}x&=\rho \cos(\theta )\sin(\phi )\\y&=\rho \sin(\theta )\sin(\phi )\\z&=\rho \cos(\phi )\end{aligned}}$ where $\displaystyle \rho ={\sqrt {x^{2}+y^{2}+z^{2}}}$. Pluging in $\displaystyle \phi ={\frac {1}{6}}\pi$ gives for $\displaystyle z$ $\displaystyle {\begin{aligned}z&=\rho \cos \left({\frac {1}{6}}\pi \right)={\frac {\sqrt {3}}{2}}\rho ={\frac {\sqrt {3}}{2}}{\sqrt {x^{2}+y^{2}+z^{2}}}\\\Rightarrow \ z^{2}&={\frac {3}{4}}(x^{2}+y^{2}+z^{2})\\\Rightarrow \ 4z^{2}&=3x^{2}+3y^{2}+3z^{2}\\\Rightarrow \ z^{2}&=3x^{2}+3y^{2}\end{aligned}}$ which is the required cone equation.

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