INSTRUCTOR: Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere x squared plus y squared plus z squared equals 4, but outside the cylinder x squared plus y squared equals 1. Let's begin with rectangular coordinates. So in rectangular coordinates, we can fairly easily see that the bounds on our x for the outer function. I'm thinking of x squared plus y squared plus z squared.
OK, boundaries on x will be negative 2 to 2, because it's a sphere with a radius of 2. We can also see that boundaries on z are going to be from negative square root of 4 minus x squared minus y squared, end square root, all the way up to square root of 4 minus x squared minus y squared, end square root. The only thing that seems odd is finding the boundaries for y for only the sphere we're talking about. But in that case, if we restrict ourselves out of the z dimension, then those boundaries become negative square root of 4 minus x squared, end square root, to square root of 4 minus x squared, end square root. That's just to find the volume of the sphere.
But in rectangular coordinates, we would take the volume of the sphere and then subtract the volume of the cylinder. So some similar logic follows. So the volume of that is between those two is two-fold.
So the integral from negative 2 to 2 with respect to x of the integral from negative square root of 4 minus x squared, end square root, to square root of 4 minus x squared, end square root, with respect to y of the integral from negative square root of 4 minus x squared minus y squared, end square root, to square root of 4 minus x squared minus y squared, end square root, with respect to z of the function 1 minus the integral from negative 1 to 1 with respect to x of the integral from negative square root of 1 minus x squared, end square root, to square root of 1 minus x squared, end square root, with respect to y of the integral from negative square root of 4 minus x squared minus y squared, and square root, to the square root of 4 minus x squared minus y squared, end squared root, with respect to z of the function 1. So that's how we would find the difference in rectangular coordinates, but the difference in volume between the outer sphere and the inner cylinder.
Now for cylindrical coordinates, you have to start thinking differently. So cylindrical coordinates will be with respect to z, r, and theta. And theta is often from 0 to 2 pi, because we are considering this in all directions.
In all directions between those two shapes, r is going to go from 1 to 2 because between those two shapes, there is a difference. There's the radius of the cylinder is 1. And the radius of the sphere is 2. So there will be our radius.
And then our z, boundaries on z, because of a relationship between z and r, we can go back and look at some of those relationships. But because of those, well, specifically, I can tell you that r squared equals x squared plus y squared. So we're replacing that into the original, or even the boundaries from the rectangular coordinates that we just wrote, even replacing it there, we see that z is going to go from negative square root of 4 minus r squared, end square root, to square root of 4 minus r squared, end square root.
So we can write this as one triple integral. And that will be the integral from 0 to 2 pi with respect to theta of the integral from 1 to 2 with respect to r of the integral from negative square root of 4 minus r squared, end square root, to square root of 4 minus r squared, end square root, with respect to z of the function r. And that will take care of the volume in cylindrical coordinates.
Now finally, we need to consider spherical coordinates. The issues are going to be, again, finding the boundaries. We need to consider phi. We need to consider rho. And we need to consider theta.
Theta will again go from 0 to 2 pi. So that one is not an issue. We do have an issue with r, or sorry, rho, in this coordinate system. There is a correspondence between r and rho, which is why I said that.
So the outer radius is rho. And that outer radius is 2. So the upper bound on rho of 2, and the inner bound isn't one, though, like it was for r because r is 1, there is a relationship.
We know that r equals rho sin phi. Since there isn't a strict boundary on rho, we need to actually use this. Since r is equal to 1, 1 equals rho multiplied by sin phi. And rearranging this, rho is equal to cosecant phi. I'm going to use that as a boundary on rho.
And then finally, finding the boundaries on z, so rho goes from cosecant phi to 2. That's our lower and upper bounds, respectively. Theta is going to go from 0 to 2 pi. But phi is another story.
So the lower and upper bounds on phi are actually the places of intersection where the sphere intersects that cylinder. So what I'm going to do is set up a small system here. I'm going to take x squared plus y squared plus z squared equals 4. And I really don't even have to write a system because the other restriction is x squared plus y equals 1.
So if I replace x squared plus y squared in this equation with 1, I have 1 plus z squared equals 4, which means that z is equal to plus or minus the square root of 3. Now, you might be wondering where I'm going with this. Why did I just find boundaries on z when I'm actually looking for values of phi.
Well, there is a relationship that we also know about. That is, phi equals arccosine, open parentheses z divided by rho, close parentheses. We have some values here, OK? We know that rho has a maximum value of 2. And we know what ours z is or some bounds on z.
So I'm going to write this in this way. Negative square root of 3 is less than or equal to z is less than or equal to the square root of 3. If I divide all sides here by 2, I'm sorry, upper bound on rho, all right, and then take the arccosine of all sides, arccosine of negative square root of 3 divided by 2 is less than or equal to phi is less than or equal to arcosine of square root of 3 over 2, close parentheses.
This gives us bounds for phi of pi over 6 is less than or equal to phi, is less than or equal to 5 pi over 6. Those are the two angle values for the intersection. Putting all this together, the spherical coordinates are the integral from pi over 6 to 5 pi over 6, with respect to phi of the integral from 0 to 2 pi with respect to theta of the integral from cosecant phi to 2 with respect to rho of the function rho squared sin of phi being our innermost function.
Now, in reality, with a lot of integrals like this, to get them set up, you want to play with all of those relationships, the relationship we have with phi and z and rho. Yeah, you want to just take those and start messing around and see if you can find some way to write each of the values in terms of things that you don't know and vise versa. And that is this volume, written in rectangular, cylindrical, and spherical coordinates.