# Integrating in cylindrical coordinates and finding surface area

I have a couple study guide involving finding integrals using cylindrical coordinates and finding one using a scalar surface. The main thing I struggle with is finding bounds.

One question asks me to express a triple integral in cylindrical coordinates for the volume of the sphere $$x^2+y^2+z^2=16$$ contained within the cylinder $$(x-2)^2+y^2=4$$.

The other questions asks me to find the surface area of the paraboloid $$z=5-2x^2-2y^2$$ that lies above the plane z=-13. The main thing I have trouble with is the bounds. Can someone explain how to do these problems for me?

I know for the first one that if the cylinder is expressed that way I can expand $$(x-2)^2$$ to $$x^2-4x+4$$ and then convert the sphere integrand and integral bounds to cylindrical. I was just asking for someone to help out because the professor did not give answers to the study guide.

For the first one, your sphere is centered at the origin with radius of $$4$$ whereas your cylinder is along $$z$$ axis, centered at $$(2, 0, 0)$$ and radius of $$2$$.

Using cylindrical coordinates,

Using $$x = r \cos \theta, y = r \sin \theta \, , \,$$your cylinder $$x^2 - 4x + y^2 = 0 \,$$ can be written as,

$$r = 4 \cos \theta \, ( - \pi/2 \leq \theta \leq \pi/2)$$. Please note the bounds of $$\theta$$.

Bounds of $$z$$ are spherical caps, $$z = \pm \sqrt{16 - x^2 - y^2} = \pm \sqrt{16 - r^2}$$

So bounds for your integral are

$$-\sqrt{16-r^2} \leq z \leq \sqrt{16-r^2}$$

$$0 \leq r \leq 4\cos \theta$$

$$-\pi/2 \leq \theta \leq \pi/2$$

For the second one,

Your parametrized surface is $$z = f(x,y) = 5 - 2x^2 - 2y^2 \,$$ which is an inverted paraboloid with vertex at $$(5, 0, 0)$$. As we need to find surface area above plane $$z = -13$$, we have

$$-13 \leq z \leq 5, 0 \leq x^2 + y^2 \leq 9$$. So the projection in $$XY$$ plane is a disc $$0 \leq r \leq 3$$.

The surface area is given by $$\iint_D \sqrt{1+f_x^2 + f_y^2} \, dA \,$$ where $$dA$$ is the projection of the surface in the $$XY$$ plane.

Taking derivative wrt $$x, y, \, f_x = -4x, f_y = -4y$$

In polar coordinates, $$dS = \sqrt{1+f_x^2 + f_y^2} \, dA = \sqrt{1 + 16r^2} \, r \, dr \, d\theta$$

Can you take it from here?

• Yes, I can. What about the other question. Also, for the first question, is the integrand just the sphere? Commented Dec 14, 2020 at 15:05
• It will be triple integral. You should have the order $dz \, dr \, d\theta$. In cylindrical coordinates $dV = r \, dz \, dr \, d\theta$. So make sure to have $r$ in your integral. Commented Dec 14, 2020 at 15:05
• Right, and the integrand will just be the sphere. Commented Dec 14, 2020 at 15:07
• I think so...don't I have to find equations of the plane and use the cross product of the vector partials? Commented Dec 14, 2020 at 16:46
• Yes you have to take partials but no vector here. That's when you have a vector field and you are doing surface integral. I will edit to show but ideally these are two separate questions and you should post separate questions. Commented Dec 14, 2020 at 16:56