A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

rotation of a bounded area question i can't solve Write the integral in one variable to find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 5.

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

    my trouble is mainly with the fact that the rotation is done around x=5 and not the x axis

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

    im sure something changes but not sure what

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

    Do you prefer the washer or shell method?

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

    well i don't know the shell method so washer i suppose

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

    Observe that the dimensions of the washer that you'll be interested in (i.e. radius) will have to be written in terms of \(y\). Let's draw the region first: |dw:1439519332453:dw|

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

    so does that mean we will have to integrate with sespect to y?

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

    Yes! Since \(y=\dfrac{1}{2}x^2\), you have \(x=\sqrt{2y}\) (taking the positive root since we're in the first quadrant). Here's a washer associated with an matching of points \((x,y)=(\sqrt{2y},y)\) and \((x,y)=(y,y)\) on the boundary of the region: |dw:1439519657597:dw|

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

    (As you'll no doubt learn soon enough, the shell method allows you to integrate with respect to \(x\) instead.)

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

    Correction: the inner radius should be \(5-\color{red}{\sqrt{2y}}\)

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

    The volume of the solid is the infinite sum of the areas of these infinitely thin washers. The area of a washer is essentially the difference of the areas of circles. Take some circle with radius \(OR\) and put some smaller circle with radius \(IR\) at the larger one's center. The area of the large circle is \(\pi(OR)^2\), and that of the smaller circle is \(\pi(IR)^2\). Remove the smaller circle from the larger, and you're left with the washer shape, which has area \(\pi((OR)^2-(IR)^2)\). So, to add up all these areas, you take the integral. It'll look something like this (I'll leave the details to you): \[V=\int_a^b \pi((OR)^2-(IR)^2)\,dy\]

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

    okay sorry lost internet connection , back now give me a second to read all that you posted

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

    okay so tell me if this is way off \[V = \pi \int\limits_{0}^{2}(y^2)-({2y}) dy\]

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

    It's off by little bit. Your setup would be right if you were revolving about the \(y\)-axis (the line \(x=0\)).

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

    okay so do i just add 5 to each expression or something?

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

    Sort of... if you take another look at that sketch, you'll notice that I lay out the lengths of a few distances. Let me remove some of that clutter: |dw:1439520860137:dw| Notice that the two smaller distances must add up to \(5\). That is, \[y+OR=5\] This means \(OR=5-y\), not just \(y\).

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

    Similarly, you can figure out what the inner radius would be by breaking down these distances: |dw:1439521024630:dw| You have \(\sqrt{2y}+IR=5\), so \(IR=5-\sqrt{2y}\).

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

    okay thats interesting let me give it another shot \[V = \pi \int\limits_{0}^{2} (-x-5)^2 - (5-\sqrt{2y}) dy\]

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

    woops forgot to square the last part

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

    I think you meant \(5-x\), or \(-x+5\) for the first term? Squared, of course.

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

    \(y\), not \(x\)**

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

    yes yes sorry about that

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

    im getting 5.333 , does that look right to you?

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

    Yup

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

    times \(\pi\), right?

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

    oh yea forgot :)

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

    It doesn't look like you actually had to do the computation, but maybe those instructions were just left out? Anyway, setting up these types of integrals is the hardest part.

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

    (usually)

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

    thanks for the help ! :D

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

    You're welcome!

  30. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.