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Idealist

  • 3 years ago

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

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  1. electrokid
    • 3 years ago
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    |dw:1363841657543:dw|

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

  3. Idealist
    • 3 years ago
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    That's not the answer, it should be 16pi*sqrt(2).

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

  6. electrokid
    • 3 years ago
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    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
    • 3 years ago
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    Could you set up the integral?

  8. electrokid
    • 3 years ago
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    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
    • 3 years ago
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    @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
    • 3 years ago
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    Thanks for your time and effort.

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