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Let's consider a two-sided an unit ice cream cone defined by $$E = \left \{ (x,y,z): x^2 + y^2 \leq z^2 \leq 1 - x^2 - y^2 \right \}$$

What is volume of this icecream cone (I only want one side because the other side is going to fall to the ground)?

Under cylindrical coordinates, the ice cream cone's volume is

$$\int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{\sqrt{1-r^2}}rdz dr d\theta $$

Now for some reason this integral gives me $0$

Under spherical coordinates, the ice cream cone's volume

$$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{1}\rho^2 \sin\phi d\rho d\phi d\theta$$

And there is no way this is $0$.

Could someone tell me what's wrong with the first integral?

Thank you

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As you know, for using the clyndrical cordinate we have to find the certain range for $r$. How it is determined? Range of $r$ would be clear if we intersect two surfaces with each other. So by solving $x^2+y^2+z^2=1$ and $z^2=x^2+y^2$ we get $2(x^2+y^2)=1$. This means that $2r^2=1$ or $0≤r≤\frac{1}{\sqrt{2}}$.

enter image description here

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  • $\begingroup$ I thought the range of r is the projection of the surface on r-theta plane? $\endgroup$ – Hawk Jul 9 '12 at 23:47
  • $\begingroup$ @jak: Yes, exactly. In $r-\theta$ plane we have a rectangle area for them. $\endgroup$ – mrs Jul 10 '12 at 4:58
  • $\begingroup$ Beautiful! Nice work! + $\endgroup$ – Namaste Mar 11 '13 at 0:07
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By using the wrong upper bound for $r$, you are adding some negative volume. You should only integrate over the range where $0 \le r \le \sqrt{1 - r^2}$, that is, from 0 to $\frac{1}{2}\sqrt{2}$.

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