Disk/Washer Method: y = x ^ 1/2 around

Disk/Washer Method

 y = x ^1/2 around the y-axis

Our mission is to find the volume that results when the curve y = x ^(1/2) between y = 0, x = 0 and x = 4 around the y-axis.

First of all draw a picture. I can't emphasize this enough.

 

We ask a few questions to organize our thoughts.

Are we integrating over x or y?

We're revolving around y, so our integral has to be in terms of y. We'll have to integrate over the y-values, 0 to 2.

One radius or two radii?

Two radii. The outer one is 4, the inner is the x value of the curve y = x ^ 1/

 

How do we set up our integral?

The outer radius is 4, so I wouldn't even bother with an integral. The "outer shape" is a cylinder of volume 32pi.

The inner radius is the x-value of the curve y = x ^ 1/2 . Since we have to integrate in terms of y, we solve for x in terms of y. Our inner radius is y^2. We integrate over the y-values, from 0 to 2.

 

The volume of the shape will be:

 

32pi -  pi times ( the integral of y^4 between y = 0 and y = 2), which is 32pi - (32pi/5) Our answer is 128pi/5.

 

Basically, what we want is the volume of a cylinder, minus the volume of a shape kind of like this: