## 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

1. electrokid Group Title

|dw:1363841657543:dw|

2. electrokid Group Title

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

3. Idealist Group Title

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

4. electrokid Group Title

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 Group Title

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

6. electrokid Group Title

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 Group Title

Could you set up the integral?

8. electrokid Group Title

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 Group Title

@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 Group Title

Thanks for your time and effort.