A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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.

  • This Question is Closed
  1. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  4. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  5. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  6. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  7. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Spin it around,|dw:1444636360587:dw|

  8. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  9. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  10. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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