# Volume that results from rotating about the X axis

• Jan 25th 2010, 05:13 PM
icepik
Volume that results from rotating about the X axis
Find the volume of the solid that results when the region enclosed by the given curves is revolved about the X-Axis:

y=(25-x^2)^(1/2)
and
y=3
• Jan 25th 2010, 05:23 PM
Lord Darkin
$
\int_3^5 \pi r^2dx
$

EDIT: r is the radius. You can also pull the pi out of the integral (as per the integral rule) to make the process of finding the integrand easier.
• Jan 25th 2010, 05:29 PM
icepik
From inspection I can see that this is a circle, but if I want to stick to just using the curve equations. It's a half circle with radius 5 and mid point 0,0. I know this.. But i want to keep it according to solids of revolution and integrate the equations ( squared ). Do i have to break it up into different segments?
• Jan 25th 2010, 05:33 PM
Lord Darkin
(Headbang) Oops, never mind. The formula I posted is if you were to rotate it around the y axis.

For x axis, I assume you would use the "washers" technique since when you rotate the solid, you'll end up with "empty space" from y=-3 to y=3.

In that case, use y=3 as the shorter radius and the other equation you have for the longer radius. (Longer radius comes first, R^2 - r^2.)
• Jan 25th 2010, 05:41 PM
icepik
from -3 to 3 under (25-x^2)^(1/2).. if rotated will that really be empty space? wouldnt that be inclusive of the empty space also?
• Jan 25th 2010, 05:48 PM
Lord Darkin
The region enclosed by the two equations:

y=(25-x^2)^(1/2)
and
y=3

When graphed in calc, looks like a semicircle above the x-axis. So when that's rotated about the x-axis, it will look like a disk with a hole in it, or a hollow cylinder. The space inside isn't included in the volume since the space from y>-3 to y<3 was not originally enclosed by the 2 functions.
• Jan 25th 2010, 05:59 PM
icepik
but how am i going to use the y co-ordinates? won't i be using the X intervals?