Rotation volume question

[akaich0u]

I recently got this problem wrong on a test, and it involves volume of a solid of rotation. For some reason I got stuck and tried to use arc length (since that was also being tested), but could someone please explain it to me?

Comments

[quantumfoam13]

As is noted on your paper, you correctly determined the total volume. You are asked to find a the plane x=k where the volume is cut in half. The set up is as follows:

(1/2)V = pi*Int[((x)^(1/2))^2,x,0,k]
=> (1/2)(8*pi) = pi*(1/2)[k^2 - 0^2]
4 = (1/2)*k^2
8 = k^2
=>k = 8^(1/2) or 2*2^(1/2)

Sorry about the format. If you'd like more explicit details, I can email them to you.

[akaich0u]

Sorry for the late reply! You made that pretty clear, actually. Thank you! :D

[where_was_i]

You could also have proceeded from the part on the paper that is circled with the c by it. The π's cancel, and the (√(x))2 becomes just x, so you're left with ∫k0xdx = ∫4kxdx. Perform the integration, and remember that k is just a number, and you'll get 1/2k2 = 1/2(42 - k2), which simplifies the same quantumfoam13 showed.

[where_was_i]

Oh, so you know why that works: you're splitting the volume into two pieces at the line x=k, and forcing the volumes of the two different shapes to be equal on the condition that k is in a particular spot. It'd be kind of like chopping a cone of ice cream in half, and trying to figure out where to cut it so it'd be equal. It looks like you understood that, but got stuck on the fact that k could be carried through the integral as a normal number would. Like I said above, k is just a normal, rational number that must live somewhere between 0 and 4 (because of the nature of the question), and you can use normal algebra to solve for it once it's in a more obvious equation.

Hope that's helpful =).

[akaich0u]

That is helpful. Thanks for your time! :D