# Thread: how to work out volume between sphere and cone?

1. ## how to work out volume between sphere and cone?

Consider a region in space bounded above by the unit sphere
x^2 + y^2 + z^2 = 1

and below by the cone z = sqrt(x^2 + y^2)

I have been trying to get my head around this. It has been so long since i have last done this stuff. am i right in thinking that I take the triple integration of one and subtract the other?

can someone give me an example of what my problem should look like? i.e. showing everything including upper and lower limits.

thanks

2. Did you forget to square the z in your equation for the cone? Assuming you did. Due to the symmetry of the region, it's easier to do this in spherical coordinates:
volume=$\displaystyle \int\int\int _V1.dV=\int_0^1\int_0^{2\pi }\int_0^{\pi /4}\rho \sin^2 (\phi ) d \phi d \theta d \rho$

3. whoops. just noticed the cone formula is: z= sqrt(x^2 + y^2). i'm sure this wil make things easier. lol

4. ojones has already told you how to do it. The sphere, $\displaystyle x^2+ y^2+ z^2= 1$, in spherical coordinates, is $\displaystyle \rho^2= 1$ and the cone, z^2= x^2+ y^2, in spherical coordinates is $\displaystyle \rho^2 cos^2(\phi)= \rho^2 sin^2(\phi)$ which reduces to $\displaystyle cos^2(\phi)= sin^2(\phi)$ or simply $\displaystyle \phi= \frac{\pi}{4}$ since $\displaystyle \phi$ must be between 0 and $\displaystyle \pi$. The volume inside the sphere, above the cone, is given by
$\displaystyle \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/4}\int_{\rho= 0}^1 \rho sin^2(\phi)d\rho d\phi d\theta$$\displaystyle = \left(\int_{\theta= 0}^{2\pi}d\theta\right)\left(\int_0^{\pi/4} sin^2(\phi)d\phi\right)\left(\int_{\rho= 0}^1 \rho d\rho\right). That is exactly what ojones told you yesterday! 5. Originally Posted by HallsofIvy ojones has already told you how to do it. The sphere, \displaystyle x^2+ y^2+ z^2= 1, in spherical coordinates, is \displaystyle \rho^2= 1 and the cone, z^2= x^2+ y^2, in spherical coordinates is \displaystyle \rho^2 cos^2(\phi)= \rho^2 sin^2(\phi) which reduces to \displaystyle cos^2(\phi)= sin^2(\phi) or simply \displaystyle \phi= \frac{\pi}{4} since \displaystyle \phi must be between 0 and \displaystyle \pi. The volume inside the sphere, above the cone, is given by \displaystyle \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/4}\int_{\rho= 0}^1 \rho sin^2(\phi)d\rho d\phi d\theta$$\displaystyle = \left(\int_{\theta= 0}^{2\pi}d\theta\right)\left(\int_0^{\pi/4} sin^2(\phi)d\phi\right)\left(\int_{\rho= 0}^1 \rho d\rho\right)$.

That is exactly what ojones told you yesterday!
I have a quick question. If I just wanted to find the volume of the sphere above the plane of intersection, how would I go about doing this? I have tried but can't figure it out. Any help would be appreciated. I have no problem solving it the way that was just shown but like to try and solve problems in more than one way. I found the volume of the cone and was just going to add the other volume to find my answer.

6. This is a little trickier. The plane in question has equation $\displaystyle z=\rho\cos\phi =\frac{1}{\sqrt 2}$. Volume will then be $\displaystyle \int_0^{2\pi}\int_0^{\pi/4}\int_{\frac{\sec \phi}{\sqrt 2}}^1\rho \sin^2\phi d\rho d\phi d\theta$.

7. Actually, if ojones had not suggesed spherical coordinates, I would have been inclined to use cylindrical coordinates for the first problem and certainly for this one. In cylindrical coordinates, the equation of the sphere is $\displaystyle r^2+ z^2= 1$ and of the cone, z= r. Those intersect where $\displaystyle r^2+ r^2= 2r^2= 1$ so at $\displaystyle z= \frac{1}{\sqrt{2}$. At that z, the equation of the sphere becomes $\displaystyle r^2+ 1/2= 1$ or $\displaystyle r^2= 1/2$, $\displaystyle r= \frac{1}{\sqrt{2}}$.

The volume under the sphere, above the plane of intersection of the cone and sphere, is $\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^{\frac{1}{\sqrt{2}}} (\sqrt{1- r^2}- \frac{1}{\sqrt{2}} r sin(\theta) drd\theta$.

8. Originally Posted by HallsofIvy
Actually, if ojones had not suggesed spherical coordinates, I would have been inclined to use cylindrical coordinates for the first problem and certainly for this one. In cylindrical coordinates, the equation of the sphere is $\displaystyle r^2+ z^2= 1$ and of the cone, z= r. Those intersect where $\displaystyle r^2+ r^2= 2r^2= 1$ so at $\displaystyle z= \frac{1}{\sqrt{2}$. At that z, the equation of the sphere becomes $\displaystyle r^2+ 1/2= 1$ or $\displaystyle r^2= 1/2$, $\displaystyle r= \frac{1}{\sqrt{2}}$.

The volume under the sphere, above the plane of intersection of the cone and sphere, is $\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^{\frac{1}{\sqrt{2}}} (\sqrt{1- r^2}- \frac{1}{\sqrt{2}} r sin(\theta) drd\theta$.
I would like to add that in computations of volume bounded by a sphere/cylinder/cone/etc I first attempt the problem using cylindrical co-ordinates and I only use spherical co-ordinates if the form that I end up with is unrecognizable.

I do this for a few reasons: the first of which is cylindrical co-ordinates is simpler in that it only has 2 bounds, theta and radius, of course we have a dz integral but this is very easily computed that we don't even need to set it up. As opposed to phi, theta and P that spherical co-ordinates introduce. Also, spherical co-ordinates involve a very clumsy dV (at least in my opinion). The second thing is that cyldrincal co-ordinates are easier to visualize and interpret and most problems involving spheres and such are easier to do with cylindrical co-ordinates!

As for generally finding the volume bounded by a sphere and some shape (a cone, a plane, a cylinder, etc) we are best served by a diagram. Of course, most of us suck as graphing 3D objects (I am horrible) but we should be able to identify rudementary shapes such as a sphere, cylinder, cone, plane, parabola, etc.

For example,

Say I want the volume bounded by

$\displaystyle z = 4 - x^2 - y^2$ and $\displaystyle z = 5+ x^2 + y^2$

Do I know how to graph that? Absolutely not, but I do know that they both involve parabolas in the XY plain (note the combination of $\displaystyle x^2 + y^2$ ) and since one is negative and one is posative, they must be parabolas interesecting eachother. We can then see that our upper z bound will be

$\displaystyle z = 5+ x^2 + y^2$

And this is all we really need to do. Of course, this type of analysis isn't good for odd looking shapes and for those I go back to the definition of the triple integral

$\displaystyle \int_a^b dx \int_{g_1 (x) }^{g_2 (x) } dy \int_{h_1 (x,y) }^{ h_2 (x,y) } dz$

We simply fit the above form in whatever way we see fit by letting x,y equal 0 at specific times and find what z function is greater on the interval. Naturally this will only work for 95% of the cases because some are actually cancelled out by symetry and if you can't draw the object you are SOL in this case. But that is a rarity and we can't really be expected to see that, because I know professors who have a hard time computing those kinds of cases and it's unlikely they would expect students to know. Unless they provide the graph to begin with!

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