1. Integration Application question

This may be hard to visualize without a graph, so I apologize in advance, but here is my question:

There is a circle with radius 'b' that is 'a' units away from the y-axis. if the circle is rotated about the y-axis, how do you find the volume of the doughnut that is formed? (assume that b<a)

I would even appreciate a jump start! Anything, really! Thank you in advanced!

2. Originally Posted by br1tt204
This may be hard to visualize without a graph, so I apologize in advance, but here is my question:

There is a circle with radius 'b' that is 'a' units away from the y-axis. if the circle is rotated about the y-axis, how do you find the volume of the doughnut that is formed? (assume that b<a)

I would even appreciate a jump start! Anything, really! Thank you in advanced!

For convinent calculation , I locate the center of the circle at

$\displaystyle (a,0)$ and thus the equation of the circle is

$\displaystyle (x-a)^2 + y^2 = b^2$

Now , let's go to the integration ,

We know the volume of the revolution about the y-axis is

$\displaystyle \pi \int_p^q [ f(y)^2 - g(y)^2 ] ~dy$ , $\displaystyle | f(y) |\geq |g(y) | , y \in [p,q]$

since $\displaystyle f(y) =$ the distance from the y -axis$\displaystyle (x)$

we should find x in terms of y , that is

$\displaystyle (x-a)^2 + y^2 = b^2 \implies x = a + \sqrt{b^2- y^2}$

and also $\displaystyle x = a - \sqrt{b^2- y^2}$

But we know $\displaystyle a + \sqrt{b^2- y^2} \geq a - \sqrt{b^2- y^2}$

For the interval , i choose $\displaystyle (-b,b)$

Now , we are able to see what the integral looks like !

It should be

$\displaystyle \pi \int_{-b}^b \left ( [ a + \sqrt{b^2- y^2} ]^2 - [ a - \sqrt{b^2- y^2}]^2 \right ) ~dy$

$\displaystyle = \pi \int_{-b}^b 4a\sqrt{b^2-y^2}~dy$

$\displaystyle = 4a\pi \int_{-b}^b \sqrt{b^2 - y^2} ~dy$

$\displaystyle = 4a \pi \times ( ~area~of ~a~semicircle ~with~radius~b ~)$

$\displaystyle = 4a \pi \cdot (\frac{\pi b^2 }{2} ) = 2ab^2\pi^2$