How can we find the volume of $y= \sqrt{16-x^2}$ rotated around

A) $y$-axis

B) $x=4$

The thing I don't understand is that the graph is a semicircle that's already on the $y$-axis. Am I supposed to rotated the whole semicircle or half of it? What do I do about part B?


For A, you can revolve either the whole semicircle or the half in the first quadrant. The resulting solid is the same. In a sense, if you revolve the whole semicircle by angle $2\pi$ you double cover the hemisphere in that if you calculate the volume swept out you will count everything twice. But I would do this by the disk method.

For part B you revolve the whole thing around $x=4$ you get a shape that is half of a doughnut with the central hole shrunk to zero. Cylindrical shells seem the way to go here.

  • $\begingroup$ for A if I'm to do the whole semicircle, Should I go from the interval -4 to 4 ? Because after the integration I'm end up with 0 on both -4 and 4, resulting 0 area, or wrong answer. $\endgroup$
    – user70994
    Apr 4 '13 at 23:20
  • $\begingroup$ for B I got the answer 64pi, am I correct ? $\endgroup$
    – user70994
    Apr 4 '13 at 23:21
  • $\begingroup$ @user70994: if you do it by disks, you are integrating along $y$ and it should go from $0$ to $4$. It would be $\int_0^4 \pi (16-y^2) dy$ If you integrate along $x$, the integrand is even so the primitive will be odd and you shouldn't get zero. Please show the integral you are doing. $\endgroup$ Apr 4 '13 at 23:25
  • $\begingroup$ My teacher wants the class to do the question with shell method, so here's what I did. Part A: $ 2pi*\int_{-4}^4 x \sqrt{16-x^2}\,dx$ Part B: $ 2pi*\int_{-4}^4 (x-4) \sqrt{16-x^2}\,dx$ $\endgroup$
    – user70994
    Apr 4 '13 at 23:32
  • $\begingroup$ For A you should only integrate from $0$ to $4$ as the shell goes all the way around and picks up the volume left of the $y$ axis. Part B looks fine. $\endgroup$ Apr 4 '13 at 23:59

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