Find the volume of the solid that lies within the sphere enter image description here, above the xy plane, and outside the cone enter image description here

My problem is finding the integral function and the limits

  • $\begingroup$ The integral function: $1,$ the limits are given by $z=\sqrt{1-x^2-y^2},$ and $z=7\sqrt{x^2+y^2}.$ $\endgroup$ – awllower Mar 27 '14 at 3:31
  • $\begingroup$ well could you give me more explanation about the limits ? $\endgroup$ – user131040 Mar 27 '14 at 3:33
  • $\begingroup$ Inside the sphere but outside the cone, so I guess it ought to look like this. No more than a comment. :D $\endgroup$ – awllower Mar 27 '14 at 3:35
  • $\begingroup$ well, I am still confused! could you please help me to get it $\endgroup$ – user131040 Mar 27 '14 at 3:37
  • $\begingroup$ Sorry, if I thought I could explain better, I would post an answer. Hope someone better than me in this area could post. In any case, thanks for responding. :) $\endgroup$ – awllower Mar 27 '14 at 3:39

First, here is a sketch showing that the cone cuts out washers at each value of $z$. The inner radius is given by the cone and the outer radius by the sphere. In cylindrical coordinates, the cone is given by $z = 7r$ and the sphere by $r^2 + z^2 = 1 \rightarrow r^2 = 1 - z^2$. The $z$ coordinate goes from $z = 0$ to whatever value of $z$ makes the radius of the cone equal to the radius of the sphere (at that $z$ value):

$$ \text{cone}: r = \frac{z}{7} \text{ plug into sphere equation}\\ \left(\frac{z}{7}\right)^2 + z^2 = 1 \rightarrow z^2 = \frac{1}{1 + \frac{1}{49}} = \frac{49}{50} \\ z_{top} = \frac{7}{\sqrt{50}} $$

Now you just need to find the volume of each differential washer:

$$ dV = Adh = \pi(R^2 - r^2)dz = \pi\left((1 - z^2) - \left(\frac{z}{7}\right)^2\right)dz $$

The outer radius, $R$, is from the sphere and the inner radius, $r$, is from the cone. So finally, you get the volume:

$$ V = \int dV = \pi\int\limits_0^{\frac{7}{\sqrt{50}}}\left(1 - \frac{50}{49}z^2\right)dz = \pi\left.\left(z - \frac{50}{147}z^3\right)\right|_0^{\frac{7}{\sqrt{50}}} \\ V = \pi \left(\frac{7}{\sqrt{50}} - \frac{50}{3\cdot49}\cdot\frac{7^3}{50\sqrt{50}}\right) \\ V = \pi\left(\frac{7}{\sqrt{50}} - \frac{7}{3\sqrt{50}}\right) = \frac{7\pi}{\sqrt{50}}\cdot \frac{2}{3} = \frac{14\pi}{3\sqrt{50}} $$


If you absolutely must, the "full" integral, in cylindrical coordinates, would be something like this:

$$ V = \int\limits_0^\frac{7}{\sqrt{50}}dz\int\limits_{\frac{z}{7}}^{\sqrt{1 - z^2}}dr\int\limits_0^{2\pi} rd\phi $$

...but it's easier to visualize it in this case and just use high school geometry to create a single integral.

  • $\begingroup$ thank you i really appreciate your help $\endgroup$ – user131040 Mar 27 '14 at 4:43

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