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Find $\large\iiint_R(x^2+y^2+z^2)\,dV$, where $R$ is the region that lies above the cone $z=c\sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2=a^2.$

I used spherical variable change to solve this problem, but the main problem is interval of $\varphi$, so I did this: $$x^2+y^2+c^2(x^2+y^2)=a^2$$ $$(c^2+1)x^2+(c^2+1)y^2=a^2$$ So it is a circle with radius $\frac{a}{\sqrt{c^2+1}}.$ Also $z$ component of intersection is: $$\frac{a^2}{c^2+1}+z^2=a^2$$ $$z=\frac{ac}{\sqrt{c^2+1}}$$

Hence, $\varphi$ interval would be from $0$ to $\arctan\left(\frac{\frac{a}{\sqrt{c^2+1}}}{\frac{ac}{\sqrt{c^2+1}}}\right)=\arctan\left(\frac{1}{c}\right)$.

But as I read solution manual, it was written $\arctan\left({\frac{1}{c}}\right)$ directly. I want to know is there a better and less complicated way to reach it?

user
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Mary Jane
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1 Answers1

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Yes, we can consider, by symmetry, a plane containing the $z$ and $x$ axes then the maximum value for $\varphi$ is the angle with respect to $z$ associated to the slope of the line $z=cx \iff x=\frac1c z$, which is indeed $\arctan({\frac{1}{c}})$.

user
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