MHB How can we solve these two volume-related questions?

[annie122]

#1 describe the solid obtained by

\pi \int_{0}^{\pi/2}\cos ^2xdx

i thought this meant that the area function is

pi \cos ^2xdx and since \cos ^2xdx = 1^2 - \sinx ^2xdx
it's the solid obtained by rotating region between y = sinx and y = 0 for 0 <= x <= pi / 2.

but this was wrong.
how?

#2 compute the volume of a torus which has inner radius R and outer radius R + r.

no idea.

 

[MarkFL]

Re: two volume-related questions

#1 describe the solid obtained by

\pi \int_{0}^{\pi/2}\cos ^2xdx

i thought this meant that the area function is

pi \cos ^2xdx and since \cos ^2xdx = 1^2 - \sinx ^2xdx
it's the solid obtained by rotating region between y = sinx and y = 0 for 0 <= x <= pi / 2.

but this was wrong.
how?

The formula for the disk method of computing the volume of a solid of revolution about the $x$-axis is:

$$V=\pi\int_a^b f^2(x)\,dx$$

Do you see now how to describe the region being rotated?

#2 compute the volume of a torus which has inner radius R and outer radius R + r.

no idea.

I would direct you to this thread which should give you an indication of how to proceed:

http://mathhelpboards.com/questions-other-sites-52/roisins-question-yahoo-answers-regarding-volume-torus-7992.html

 

[annie122]

Re: two volume-related questions

i think i now get #1but i don't know how your link relates to my question.
is the solid in question in the thread a torus??

 

[MarkFL]

Re: two volume-related questions

...
but i don't know how your link relates to my question.
is the solid in question in the thread a torus??

Yes, the names of the variables are different, but it is a torus. I just wanted you to see the method used.

To work the second problem using the variables given, I would recommend revolving a circle of radius $\dfrac{r}{2}$ and centered at $\left(R+\dfrac{r}{2},0 \right)$ about the $y$-axis. Do you see how this will create a torus whose inner radius is $R$ and whose outer radius is $R+r$?