Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Andresfon12

  • 3 years ago

evaluate the triple integral

  • This Question is Closed
  1. Andresfon12
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits_{0}^{5} \int\limits_{0}^{1} \int\limits_{0}^{\sqrt {1-z^2}} ze^y dx dz dy\]

  2. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what's the problem you're having?

  3. Andresfon12
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    inter sqrt 0_1-z^2 and finding dx

  4. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int dx=x\]so\[\int_0^5\int_0^1\int_0^{\sqrt{1-z^2}}ze^ydxdzdy=\int_0^5\int_0^1\left.ze^yx\right]_0^{\sqrt{1-z^2}}dzdy\]\[=\int_0^5\int_0^1ze^y\sqrt{1-z^2}dzdy\]and keep working from the inside out...

  5. Andresfon12
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that mean \[\int\limits_{0}^{5} \int\limits_{0}^{1} z^2/2e^y *2/3 (1-z^2) z ? dy\]

  6. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    all the parts not dependent on z can be taken out of the inner integral and treated as constants\[\int_0^5\int_0^1ze^y\sqrt{1-z^2}dzdy=\int_0^5e^y\left[\int_0^1z\sqrt{1-z^2}dz\right]dy\]what is this inner integral?

  7. Andresfon12
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i guess i can substitution let u= 1-z^2 du=2z dz => du/2 =z dz

  8. Andresfon12
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    use*

  9. 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