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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.

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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?

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  • $\begingroup$ Yes, I can. What about the other question. Also, for the first question, is the integrand just the sphere? $\endgroup$ Commented Dec 14, 2020 at 15:05
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    $\begingroup$ 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. $\endgroup$
    – Math Lover
    Commented Dec 14, 2020 at 15:05
  • $\begingroup$ Right, and the integrand will just be the sphere. $\endgroup$ Commented Dec 14, 2020 at 15:07
  • $\begingroup$ I think so...don't I have to find equations of the plane and use the cross product of the vector partials? $\endgroup$ Commented Dec 14, 2020 at 16:46
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    $\begingroup$ 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. $\endgroup$
    – Math Lover
    Commented Dec 14, 2020 at 16:56

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