Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

Could you set up the integral?

Answer 7

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

Answer 8

@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