Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Idealist

  • 2 years ago

Compute the surface area of the portion of the cone z=sqrt(x^2+y^2) below the plane z=4.

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

    |dw:1363841657543:dw|

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

    \[ dA=2\pi\,r(z)\,dz\\ r = z\\ A=2\pi\int_0^4zdz=2\pi\left[{z^2\over 2}\right]_0^4=16\pi \]

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

    That's not the answer, it should be 16pi*sqrt(2).

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

    my bad.. \[ A=\iint_z\sqrt{\left({\partial z\over \partial x}\right)^2+\left({\partial z\over \partial y}\right)^2+1}dx\,dy \]

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

    to get the limits, \[z=4\\x^2+y^2=(2)^2\] circle of radius 2 so, \(0\le x,y\le2\)

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

    Method 2: Polar co-ordinates Use \(x=r\cos\theta\\y=r\sin\theta\) \[z(r,\theta)=r^2\\0\le r\le2,\qquad 0\le\theta\le2\pi\]

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

    Could you set up the integral?

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

    correction: \[A=\iint_R\sqrt{\ldots}(\pi\,dx\,dy)\] so, \[A=\int_{x=-4}^4\int_{y=0}^2\sqrt{{x^2\over x^2+y^2}+{y^2\over x^2+y^2}+1}dx\,dy\\ A=\sqrt{2}\int_{x=-4}^4\left(\int_{y=-\sqrt{16-x^2}}^\sqrt{16-x^2}dy\right)dx\\ A=\sqrt{2}\int_{-4}^42\sqrt{16-x^2}dx\\ \text{use }\;x=4\sin\theta\implies\theta=\sin^{-1}(x/4)\implies dx=4\cos\theta d\theta\\ \text{when}\;x=-4,\quad\theta=-{\pi\over2}\\ \text{when}\;x=4,\quad\theta={\pi\over2}\\ A=2\sqrt{2}\int_{-\pi\over2}^{\pi\over2}2\cos\theta(4\cos\theta d\theta)\\ A=8\sqrt{2}\int_{-\pi\over2}^{\pi\over2}(1-\cos2\theta)d\theta A=8\sqrt{2}\left[\theta-{\sin2\theta\over2}\right]_{-\pi\over2}^{\pi\over2}\\ A=8\sqrt{2}\left[{\pi\over2}+{\pi\over2}\right]\\ \boxed{A=8\pi\sqrt{2}} \] @satellite73 could you check plz

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

    @Idealist This is Half of the surface of the cone rotating one way from x=-4 to x=4 the the other half from x=4 to x=-4

  10. Idealist
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks for your time and effort.

  11. 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.