anonymous
  • anonymous
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.
Mathematics
jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
"...using discs/washers" it's way easier just using calculus have you had a go?
anonymous
  • 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 : \[\pi \int\limits_{0}^{2}(\sqrt{y-1}^2-(0)^2dy\]
zepdrix
  • zepdrix
Hmm :) Ya if we look at the graph...

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zepdrix
  • zepdrix
|dw:1444636132588:dw|Notice that down here, your left boundary is NOT the function x!
zepdrix
  • zepdrix
So when you're doing this problem, where the boundaries change, you want to split it into two separate integrals.
zepdrix
  • zepdrix
|dw:1444636317000:dw|I guess you don't really need an integral for this lower part here.
zepdrix
  • zepdrix
Spin it around,|dw:1444636360587:dw|
zepdrix
  • zepdrix
volume of that cylinder = (surface area)(height)\[\large\rm v=(\pi\cdot1^2)(1)=\pi\]
zepdrix
  • zepdrix
And your integral looks a little goofed up.. hmmm thinking...
zepdrix
  • 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.
zepdrix
  • 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.
zepdrix
  • zepdrix
|dw:1444636696922:dw|This will be our \(\large\rm \color{blue}{R}\) and \(\large\rm \color{red}{r}\).