## anonymous one year ago Find the volume of the solid that results when the region bounded by y=0, x=0, x=1, and y=x^2+1 is revolved around the y-axis using discs/washers.

1. IrishBoy123

"...using discs/washers" it's way easier just using calculus have you had a go?

2. anonymous

I know how to use disks/washers for most problems, but after I graphed this one, I couldn't figure it the equation. It's wrong, but I did this : $\pi \int\limits_{0}^{2}(\sqrt{y-1}^2-(0)^2dy$

3. zepdrix

Hmm :) Ya if we look at the graph...

4. zepdrix

|dw:1444636132588:dw|Notice that down here, your left boundary is NOT the function x!

5. zepdrix

So when you're doing this problem, where the boundaries change, you want to split it into two separate integrals.

6. zepdrix

|dw:1444636317000:dw|I guess you don't really need an integral for this lower part here.

7. zepdrix

Spin it around,|dw:1444636360587:dw|

8. zepdrix

volume of that cylinder = (surface area)(height)$\large\rm v=(\pi\cdot1^2)(1)=\pi$

9. zepdrix

And your integral looks a little goofed up.. hmmm thinking...

10. zepdrix

|dw:1444636534244:dw|So we've figured out this bottom section here. Let's take a slice vertically of this upper region with "thickness" dy.

11. zepdrix

We get a washer, which is a disk with a hole in it. So you have the right idea there. But you're subtracting incorrectly. Hmm, let's see if we can fix that.

12. zepdrix

|dw:1444636696922:dw|This will be our $$\large\rm \color{blue}{R}$$ and $$\large\rm \color{red}{r}$$.