Science:Math Exam Resources/Courses/MATH200/December 2011/Question 08 (b) i/Solution 1

Region (1) is part of head or body?

In the integral (1), spherical coordinates are used, where the radius covers the interval $\displaystyle \rho \in [0,2]$, the horizontal angle covers $\displaystyle \theta \in [0,2\pi ]$ which is the whole circle and the vertical angle covers $\displaystyle \phi \in [0,2/3\pi ]$. Hence, the region of integration is a ball with radius $\displaystyle 2$, where a cone is cut out from below.

Since the head of the snowman is a ball with radius $\displaystyle 2$, this region must be part of the head.

Finding the equation of the enclosing sphere

Since the region of integration is part of the ball with radius $\displaystyle 2$, whose centre is in the origin, it is enclosed by the sphere $\displaystyle x^{2}+y^{2}+z^{2}=2^{2}=4$.

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 $\displaystyle \phi$, as seen in the following figure, which shows a cross-section of the region of integration through the $\displaystyle zy$-plane.

To find the equation of the cone, we must rewrite the expression $\displaystyle \phi ={\frac {2}{3}}\pi$ 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 $\displaystyle \phi ={\frac {2}{3}}\pi$ gives for $\displaystyle z$

$\displaystyle {\begin{aligned}z&=\rho \cos \left({\frac {2}{3}}\pi \right)=-{\frac {1}{2}}\rho =-{\frac {1}{2}}{\sqrt {x^{2}+y^{2}+z^{2}}}\\\Rightarrow \ z^{2}&={\frac {1}{4}}(x^{2}+y^{2}+z^{2})\\\Rightarrow \ 4z^{2}&=x^{2}+y^{2}+z^{2}\\\Rightarrow \ 3z^{2}&=x^{2}+y^{2}\end{aligned}}$

which is the required cone equation.