# Thread: I need tips and tricks for setting up cylindrical and spherical triple integrals.

1. ## I need tips and tricks for setting up cylindrical and spherical triple integrals.

Ok, so solving triple integrals in Cartesian, cylindrical, and spherical is not a problem for me. I rip through solving them no problem. My issue is setting them up. When given a lower bound and an upper bound (normally always given in Cartesian - x,y,z - form) I always seem to screw one of them up, which means I screw up the whole problem. Any tips or tricks to help me see why we use the bounds we do for cylindrical and spherical problems would be greatly appreciated!

Some example problems - all solving for the volume of some region. (I just need the integrals set up and why not solved):

1.) Find the volume of the region E bounded by the paraboloids z = x^2 + y^2 and z = 36 - 8x^2 - 8y^2. Use cylindrical coordinates.

2.) Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 49, above the xy-plane, and below the following cone.

3.) Evaluate the integral below, where E lies between the spheres x^2 + y^2 + z^2 = 1 and x^2 + y^2 + z^2 = 4 in the first octant.

Any rules of thumb I should follow all the time etc would be greatly appreciated. Usually solving for r or p, depending on if its cylindrical or spherical, is never a problem. But normally at least one of the integrals is 0 - 2pi but I can't seem to find a reasoning as to why...

2. For 1, when I learned multiple integration I found useful to think in a way that the coordinate system gives you parameters that describes the solid, or the region you are dealing with. Although this could mean the same as we always do, maybe it helps you to understand why the integrals vary from 0 to $\displaystyle 2\pi$, for example.

So imagine the solid's shadow in the xy plane: you will get a big circunference with radius 2 formed by the intersection of the two paraboloids. (just plug one z into the other one, then you will find the equation of that circunference I'm telling you). We must have $\displaystyle 0\leq r\leq 2$ - in a heuristics way, we will sum a series of solids each one with different radius, but similiar in the shape.

For the lower and upper bound of z (the height) you should see that $\displaystyle z=x^2+y^2$ gives you the lowest heigth and $\displaystyle z=36-8x^2-8y^2$ gives you the highest height, but, every size will depends on the value of r so we must write the two equations in polar form and we will use the r as the limits for the integral with respect of z, or the height. If $\displaystyle z=x^2+y^2$, then by doing $\displaystyle x=r\cos t$ and $\displaystyle y=r\sin t$ we obtain $\displaystyle z=r^2$ and for the other $\displaystyle z=36-8x^2-8y^2$, we must have $\displaystyle z=36-8(x^2+y^2) = 36 - 6r^2$ by the same substitution.

Finally, fix a radius, and fix a maximum and minum height - you should have a line describing the contour of the solid. If you rotate this contour by 360 degrees, or $\displaystyle 2\pi$, you will obtain a surface of revolution. Thats why the angular parameter of the cylindrical coordinates should vary from 0 to $\displaystyle 2\pi$.

So mix all these together and think a little about it then you should see that this makes sense in a weird way - we will find the volume of the solid adding all these "shells" that grow in size as we change r. But if you see, we must integrate two times to get a "thin plate", then the rotation of this will give us the required solid.

So, we must write the integral taking care that the order of integration is correct, because we see that the height depends on r, so we must integrate first the height then the r:

$\displaystyle \int\int\int dV = \int\int\int_{r}^{36-8r} dz dA$
$\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\int_{r}^{36-8r}~dz~r~dr~dt$

Noticing that $\displaystyle dA$ is the area element in the xy plane which is $\displaystyle dA=r~dr~dt$.

I hope this helps you in this exercise and in the others and I apologize for any mistake in the english.

3. But it isn't ALWAYS 0 - 2pi. For instance, if you want to solve for the region that lies between the sphere x^2 + y^2 + z^2 = 4 and cylinder x^2 + y^2 -2x = 0, then the bounds (in spherical terms) for phi and theta are both from 0 -> pi/2 and p is from 0 -> 2. So why isn't this from 0 -> 2pi?? Usually solving for the 'height' isn't my issue. I just keep getting caught up on the (x,y) or (phi,theta) in spherical or (theta,r) in cylindrical - as my book puts it.

4. It will really help if you draw them out. The sphere is centered at the origin with radius 2. The cylinder need a bit of work to make it right.

$\displaystyle x^2 + y^2 -2x = 0$
$\displaystyle x^2-2x+1-1+y^2=0$
$\displaystyle (x-1)^2+y^2=1$

You can see that it's a cyilnder centered on (1,0) in the xy plane and we are letting z go freely. The radius is just 1.They intersect only on one side of the origin and hence you don't need the full 2 $\displaystyle \pi$.

Soo, phi should actually go from 0 to pi because the cylinder extends above and below the xy plane. For me, phi is the angle between the position vector and the positive z axis. However, due to the symmetry of the problem the whole volume is twice the volume if you just go from $\displaystyle \phi = 0$ to $\displaystyle \pi/2$.

Now, for theta, there is also a symmetry you can exploit. Theta needs to go something like from 0 to $\displaystyle \pi$, because the cylinder is only on one side of the yz plane. But the integral from 0 to $\displaystyle \pi$ is just twice the integral from 0 to $\displaystyle \pi/2$

So the integral is from $\displaystyle \phi = 0$ to $\displaystyle \pi$ and $\displaystyle \theta =0$ to $\displaystyle \pi$, which is equal to 4 times the integral from $\displaystyle \phi = 0$ to $\displaystyle \pi/2$ and $\displaystyle \theta = 0$ to $\displaystyle \pi/2$

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# tips for spherical coordinate integrals

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