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I really have a hard time understanding how to solve problems where you have two surfaces intersecting. The problem I'm stuck on now is the following:

Find the area of the part of the sphere $x^2+y^2+z^2=4a^2$ that lies inside the cylinder $x^2+y^2=2ay$.

This problem is stated in a chapter about surface integrals, so I must find the area element $dS$. But I don't understand how to find it, or where to start. The book has a solution to this problem, but I don't really understand what they do (see pictures below). Like what do they do in this when writing $2z (dz/dy)=-2x$ ?

enter image description here enter image description here

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  • $\begingroup$ Take the partial derivative $\partial x$ on the whole thing: $x^2+y^2+z^2=4a^2$ we get $2x + 0 + 2z \cdot \partial z/ \partial x = 0$. The reason $\partial y / \partial x=0$ is because the two coordinates are independent (in this context $y$ is "free" to be anything given a $x$ value), and the reason $\partial z / \partial x$ is not zero is because $z = z(x,y)$ is a surface under consideration. $\endgroup$ Nov 18, 2021 at 20:48
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    $\begingroup$ Do you have other questions regarding your solution from the book? Please list all of them, otherwise people will have to make blind guesses when composing an answer (this is partially why I'm commenting instead of making a post). $\endgroup$ Nov 18, 2021 at 20:49
  • $\begingroup$ @LeeDavidChungLin I wanted to ask only one question first to see if I could understand the rest of the solution myself. I understand now what you mean by people having to make a blind guess, and I will remember that the next time I post a question! Thank you for your reply $\endgroup$
    – Mathomat55
    Nov 19, 2021 at 22:51
  • $\begingroup$ I understand. People have to meet others halfway. In this particular case, I was too lazy to type up a full-blown answer (not specific to your needs) while someone else was willing, and this is what makes this site work (that there is diversity in assessment of the situations). $\endgroup$ Nov 19, 2021 at 23:02

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I've taken a couple of classes that cover surface integration, and it is often explained in a couple of different ways. I'll try and lay out what made most sense to me. \begin{equation} \iint_\mathbb{R}\,dS = \iint_\mathbb{R}\|\mathbf{n}\|\,dudv \end{equation}

where $\mathbf{n}$ is a special normal vector. One way that we often find this vector is by solving for $z$ in terms of $x$ and $y$, so that we can parameterize the surface with $u = x$ and $v = y$. Because we make this substitution, our integral turns into: \begin{equation} \iint_\mathbb{R}\|\mathbf{n}\|\,dxdy \end{equation} Now we have to solve for $z$ in terms of $x$ and $y$. \begin{equation} \mathbf{r}(x,y) = \begin{bmatrix} x \\ y \\ \sqrt{4a^2 - x^2 - y^2} \end{bmatrix} \end{equation} Just a note, notice that for $-2a \lt x \lt 2a$ and $-2a \lt y \lt 2a$, plugging in any $x$, $y$ pair to $\mathbf{r}(x,y)$ will give you a point on this sphere! That's what a surface parameterization does. Alright, so, now that we have $\mathbf{r}(x,y)$, we need to compute $\mathbf{n}$, which is defined as:

\begin{equation} \mathbf{n} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial (x)}{\partial x} & \frac{\partial (y)}{\partial x} & \frac{\partial (\sqrt{4a^2 - x^2 - y^2})}{\partial x} \\ \frac{\partial (x)}{\partial y} & \frac{\partial (y)}{\partial y} & \frac{\partial (\sqrt{4a^2 - x^2 - y^2})}{\partial y} \end{bmatrix} \end{equation} \begin{equation} \frac{\partial \:}{\partial \:x}\left(\sqrt{4a^2-x^2-y^2}\right) = \frac{1}{2\sqrt{4a^2-x^2-y^2}}\left(-2x\right) = -\frac{x}{\sqrt{4a^2-x^2-y^2}} = -\frac{x}{z} \end{equation} \begin{equation} \frac{\partial \:}{\partial \:y}\left(\sqrt{4a^2-x^2-y^2}\right) = \frac{1}{2\sqrt{4a^2-x^2-y^2}}\left(-2y\right) = -\frac{y}{\sqrt{4a^2-x^2-y^2}} = \frac{-y}{z} \end{equation} \begin{equation} \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -\frac{x}{z} \\ 0 & 1 & -\frac{y}{z} \end{bmatrix} = \begin{bmatrix} \frac{x}{z} \\ \frac{x}{y} \\ 1 \end{bmatrix} \end{equation} Since we need the magnitude, you don't have to worry about the sign, but in general this normal vector should always point outward from the surface. \begin{equation} \|\mathbf{n}\| = \sqrt{(\frac{x}{z})^2 + (\frac{y}{z})^2 + (1)^2} = \sqrt{1 + \frac{x^2 + y^2}{z^2}} \end{equation} Finally! We have the equation that is listed in your textbook for $dS$. The integral becomes: \begin{equation} \iint_\mathbb{R} \sqrt{1 + \frac{x^2 + y^2}{z^2}} \,dxdy \end{equation} Now just substitute in the equation for $z$ and we get the integral that they convert to polar coordinates. There's a lot of steps in the solution they provide that seem to come out of thin air. Hopefully this helps clear it up.

EDIT to answer:

But here I don't understand using dxdy works when the area does not even lay in the xy-plane. I just really want to visualize how this works.

One really helpful thing I was taught is to think of surface integration as projecting the surface area that you want to compute onto the plane that you want to integrate over. How does this end up giving you the surface area of the region that you are projecting? Through $\|\mathbf{n}\|$. Adding this to the inside of the integral scales every little piece of the integration in just the right way so that our normal area integration in the x-y plane actually gives us the surface area of the region above it. Here's a quick sketch: enter image description here

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  • $\begingroup$ Wow! Thank you so much! I have these equations in my book, but solution, I did not even realize that they used the them. Your parametrization represents the whole sphere. I have a hard time visualizing how the bonds $0 \leq r \leq 2a sin \theta$ and $0 \leq \theta \leq \pi/2$ gives us the area. I know that the bonds come from the cylinder. But I'm used to triple integrals where on can imagine summing up the area of infinite layers to get the volume. But here I don't understand using dxdy works when the area does not even lay in the xy-plane. I just really want to visualize how this works. $\endgroup$
    – Mathomat55
    Nov 19, 2021 at 10:03
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    $\begingroup$ Yes, so my parameterization represents the top half of the sphere (the part that lies above the x-y plane), and we're computing the surface area of that portion above the half circle they have drawn in the x-y plane. How does integrating over an area in the x-y plane give us the surface area of the partial sphere above it? I'll add to my answer because it's actually pretty cool. $\endgroup$ Nov 19, 2021 at 11:56
  • $\begingroup$ You made my day, and you are really good at explaining this! I really appreciate all your help :) $\endgroup$
    – Mathomat55
    Nov 19, 2021 at 22:48

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