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Idealist
Group Title
Compute the surface area of the portion of the cone z=sqrt(x^2+y^2) below the plane z=4.
 one year ago
 one year ago
Idealist Group Title
Compute the surface area of the portion of the cone z=sqrt(x^2+y^2) below the plane z=4.
 one year ago
 one year ago

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electrokid Group TitleBest ResponseYou've already chosen the best response.1
dw:1363841657543:dw
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
\[ dA=2\pi\,r(z)\,dz\\ r = z\\ A=2\pi\int_0^4zdz=2\pi\left[{z^2\over 2}\right]_0^4=16\pi \]
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
That's not the answer, it should be 16pi*sqrt(2).
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
my bad.. \[ A=\iint_z\sqrt{\left({\partial z\over \partial x}\right)^2+\left({\partial z\over \partial y}\right)^2+1}dx\,dy \]
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
to get the limits, \[z=4\\x^2+y^2=(2)^2\] circle of radius 2 so, \(0\le x,y\le2\)
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
Method 2: Polar coordinates Use \(x=r\cos\theta\\y=r\sin\theta\) \[z(r,\theta)=r^2\\0\le r\le2,\qquad 0\le\theta\le2\pi\]
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
Could you set up the integral?
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
correction: \[A=\iint_R\sqrt{\ldots}(\pi\,dx\,dy)\] so, \[A=\int_{x=4}^4\int_{y=0}^2\sqrt{{x^2\over x^2+y^2}+{y^2\over x^2+y^2}+1}dx\,dy\\ A=\sqrt{2}\int_{x=4}^4\left(\int_{y=\sqrt{16x^2}}^\sqrt{16x^2}dy\right)dx\\ A=\sqrt{2}\int_{4}^42\sqrt{16x^2}dx\\ \text{use }\;x=4\sin\theta\implies\theta=\sin^{1}(x/4)\implies dx=4\cos\theta d\theta\\ \text{when}\;x=4,\quad\theta={\pi\over2}\\ \text{when}\;x=4,\quad\theta={\pi\over2}\\ A=2\sqrt{2}\int_{\pi\over2}^{\pi\over2}2\cos\theta(4\cos\theta d\theta)\\ A=8\sqrt{2}\int_{\pi\over2}^{\pi\over2}(1\cos2\theta)d\theta A=8\sqrt{2}\left[\theta{\sin2\theta\over2}\right]_{\pi\over2}^{\pi\over2}\\ A=8\sqrt{2}\left[{\pi\over2}+{\pi\over2}\right]\\ \boxed{A=8\pi\sqrt{2}} \] @satellite73 could you check plz
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
Thanks for your time and effort.
 one year ago
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