Question

Why are the bounds from 0 to 2 on 4G-5? The question states, "Find the area of y=x^2, where x is between 0 and 4, revolved around the y axis." Shouldn't the bounds be 0 to 16 when integrating with respect to y since that's the range of y? Also, why does the solution use the negative root for the derivative when it seems like we should only care about the positive root?

Answer 1

Could you provide a link to this problem?

Answer 2

My only way of interpreting this is: \( \int_{0}^{4}x^2 dx=\frac{64}{3}\).

Answer 3

It's a revolution problem, not straightforward integration.4G-5

Problem: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-second-fundamental-theorem-areas-volumes/problem-set-7/MIT18_01SC_pset4prb.pdf

Solution: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-second-fundamental-theorem-areas-volumes/problem-set-7/MIT18_01SC_pset4sol.pdf

Answer 4

You end up integrating \[∫2π \sqrt(y+1/4)dy\] The solutions manual says it should be from 0 to 2, but given that
y=x2
shouldn't the bounds be from 0 to 16? Aren't you supposed to plug in the range of x into the equation in order to get the bounds for y?

Answer 5

I have shown this question to others and I myself cannot come to an answer. I don't know precisely what makes \([0,2]\) the proper interval of integration.

Answer 6

The solution is in error. The limits should be from 0 to 16.
The problem can also be done by integrating\[\int\limits_{0}^{4}2pix \sqrt{1+4x^{2}}dx\] Either way gets the correct answer.

Answer 7

@Stacey, really?! I was worried that, for once, I literally did not understand something despite putting a lot of focus and attention in it. I looked at just about every source I could and couldn't come to a different conclusion. But, I was thinking that M.I.T. professors wouldn't be so careless. Oh well, everyone's human. :)