# Homework Help: Finding the surface area of one volume contained within another

1. Dec 17, 2011

### tjackson3

1. The problem statement, all variables and given/known data

This isn't actually a homework problem, but rather a class of problems I'm running into as I study for prelims. I'm taking these from Greenspan's Calculus: An Introduction to Applied Mathematics. This type of problem has come up in the context of both volume and surface area, but I'm hoping figuring out the surface area one will be sufficient to figure out the volume one. Here's an example:

Find the surface area of the sphere $x^2+y^2+z^2 = a^2$ contained within the cylinder $x^2+y^2 = ax.$

2. Relevant equations

Surface area can be determined using a surface integral, $\iint_S\ dS$, and in this case, $dS = a\sin\theta\ d\theta d\phi$

3. The attempt at a solution

My original thought was to set the two equations equal to each other to get an expression for z. This results in $z = \pm\sqrt{a^2-ax}$. I've seen one strategy where you put this into the surface integral equation instead of just dS, though I don't understand why, and even if you did, what would the limits of integration be? Just 0 to 2$\pi$ and 0 to $\pi[/itx]? Thanks! 2. Dec 17, 2011 ### Frogeyedpeas The cylinder x^2 + y^2 = ax is the solution to the equation x/a +y^2/(ax) = 0 which isn't a cylinder at all... In fact its more similar to a parabolic-conic like object, that is nested in a plane and is undefined at zero... 3. Dec 17, 2011 ### tjackson3 It is defined at zero. When you divided through by x, you forgot to consider the case when x=0. There the surface is equal to zero, making it conic as you suggest. I'm just using the author's words though - he does seem to throw around the term cylinder a lot 4. Dec 17, 2011 ### Frogeyedpeas So what is the problem suggesting? depending on how much of the graph is defined there can be successively larger spheres inscribed within the conic... Actually I take that statement back because we still have another problem: The sphere x^2 + y^2 +z^2 = a^2 is not even completely contained in the conic... Does he mean the surface area that is contained within the conic because that should be able to be solved for Last edited: Dec 17, 2011 5. Dec 17, 2011 ### tjackson3 Yes, that's what he meant. 6. Dec 18, 2011 ### tjackson3 Here's an alternate example from Stewart that might be easier: Find the mass of the lamina with constant density function δ where the lamina is the portion of the circular cylinder [itex]x^2+z^2=4$ that is above the rectangle R, where $0 \leq x \leq 1, 0 \leq y \leq 4$

7. Dec 18, 2011

### LCKurtz

If you complete the square on $x^2-ax+y^2=0$ you will find it is a circular cylinder of radius a/2 centered at (a/2,0), which intersects the sphere in two pieces, above and below the xy plane. That equation has a particularly nice equation in terms of polar coordinates $(r,\theta)$. So I would suggest you express the sphere parametrically as $\vec R(r,θ)$. Use the parametric form of dS and the polar $(r,\theta)$ limits.