Using disk method with respect to y-axis

[Geekchick]

Homework Statement

y=$$\sqrt{x}$$, x=4, y=0

Homework Equations

$$\pi$$$$\int^{c}_{d}{R(y)}^{2}$$dy

The Attempt at a Solution

I solve for x then pluged it into the formula

$$\int$$$$^{2}_{0}$${(y$$^{2}$$)$$^{2}$$}

The answer I got was $$\frac{32\pi}{5}$$

but my textbook says the answer should be $$\frac{128\pi}{5}$$

 

[Dick]

It looks like what they want is the area between y=sqrt(x) and the vertical line x=4 rotated around the y-axis. You'll want to use washers. Or you could subtract your answer from the volume of the cylinder with radius 4 and height 2.

 

[Geekchick]

Yeah sorry, I thought I put that I was supposed to find the the volume as it rotates around the y-axis. which is what i tried to do by solving y=sqrt(x) for x=y^2 then I used that boundries 0-2 and integrated. How do you suppose I use washers? I thought the problem had to have two functions of x to do that.

Please help this problem has been bothering me since yesterday!
Thanks!

 

[Dick]

If you cut the volume at a fixed value of y then you get a washer shaped region (a circle with a circular hole in the middle). The inner radius is where it hits y=sqrt(x) and, as you said, that's y^2. The outer radius is where it hits the line x=4. So the outer radius is 4. To use the method of washers, you integrate pi*(outer radius)^2-pi*(inner radius)^2.

 

[Geekchick]

I got it! thank you so much! I can always depend on the PF members!

 

[Geekchick]

Oh, I was wondering can the disk method ever be used to find the volume around the y-axis? If so how can I tell? thanks!

 

[Dick]

Draw a sketch of the region. Do you see what you were doing wrong the first time? You were rotating the region between x=0 (the y-axis) and y=sqrt(x). Disks work fine for that. There's no hole in the middle. But what they wanted was the region between y=sqrt(x) and x=4. Now there's a hole in it and you should use washers. You were rotating the wrong region.

 

[Geekchick]

Ok, so for the volume of the same function about the x-axis the graph essentialliy swings down and around the axis with no gap in between the values of integation 0-4. However, by trying to find the volume of the same function as it swings around the y-axis there is a gap between the values of integration 0-2 which is accounted for by the funtion x=y^2. Right?

 

[Dick]

That sounds right. Because in your problem they wanted you to rotate the region OUTSIDE of the function, not inside.