2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.