Calculus II problem: Volume of the solid

• Jul 5th 2009, 10:42 PM
kn336a
Calculus II problem: Volume of the solid
Hi, I have one problem that I am completely stuck on. It is

x^2 + y^2 - 6x + 8 = 0

The problem I have is setting up the actual equation. Do I make everything in terms of y, then find where y = 0 for the points?

I tried doing this and it gave me a really funky looking graph. If anybody could point me in the right direction, that would be great!(Hi)

Thanks!
• Jul 5th 2009, 10:59 PM
earboth
Quote:

Originally Posted by kn336a
Hi, I have one problem that I am completely stuck on. It is

x^2 + y^2 - 6x + 8 = 0

The problem I have is setting up the actual equation. Do I make everything in terms of y, then find where y = 0 for the points?

I tried doing this and it gave me a really funky looking graph. If anybody could point me in the right direction, that would be great!(Hi)

Thanks!

Complete the squares:

$x^2 - 6x \bold{\color{red}+9}+ y^2 =- 8 \bold{\color{red}+9}$

$(x-3)^2+y^2=1$

This equation describes a circle with the center at C(3,0) and the radius r = 1
• Jul 6th 2009, 12:39 AM
kn336a
If I am trying to integrate that, I just have to put that in terms of dy or dx right?
• Jul 6th 2009, 01:04 AM
matheagle
are you trying to find volume or area?
as pointed out this is a circle in just two dimensions.
are you revolving it about a line or is there some constraints on the z-axis?
• Jul 6th 2009, 01:27 AM
kn336a
Sorry, here is my entire question.

Find the volume of the solid obtained by revolving the region bounded by the curve x^2 + y^2 - 6x + 8 = 0

im pretty stuck right now.
• Jul 6th 2009, 01:43 AM
matheagle
Quote:

Originally Posted by kn336a
Sorry, here is my entire question.

Find the volume of the solid obtained by revolving the region bounded by the curve x^2 + y^2 - 6x + 8 = 0

im pretty stuck right now.

still incomplete