How to find the radius in each of these integrals?

[Celestor]

Homework Statement

Given: y = sqrt(x), y=0, x=3

Find the volume of the solid bounded by these functions, revolved around B) y-axis and C) line x=3. (Disk method)

Homework Equations

Disk method of finding volume using π ∫ r²dy

The Attempt at a Solution

Ok so, the part of the problem that I am having trouble on is how to find the radius. I know that both integrals will be with respect to dy with the limits being 0 and sqrt(3) and I understand that these two integrals SHOULD be different, but I'm not sure what that difference is in terms of the radius in πr².

Please refer to the solution to B and C given in this link (http://www.calcchat.com/book/Calculus-10e/7/2/11/). My question is, why is the radius [3² - (y²)²] in B and why is the radius [3-y²]² in C. My problem is understanding why the power of 2 is distributed differently and how I'm supposed to interpret that based on the problem.

Thanks,

 

[BvU]

Hello Celestor, welcome to PF !

Your question is a bit strange. The radius itself isn't [ 3²-(y²)² ] in b. What do you get when you rotate the shaded area around the y-axis ?

[edit] I mean the dark shaded rectangular area...​

In (c) you rotate around the line x = 3, so what do you get when you rotate the shaded area around the dashed line ?

[edit] cute that you get to see the exercise worked out at the website. But doesn't that prevent you from first trying to find your own path towards a solution ?​