# Math Help - Polar Co-ordinates

1. ## Polar Co-ordinates

Find the surface area of the circular paraboloid
$z = x^2+ y^2" alt="z = x^2+ y^2" />between

the
xy-plane and the plane z = 4.

The limits have suddeny become for double integration

2pi 0
and 3 and 0
dr dtheta

I know

r = (x^2 + y^2)^0.5
x = r cos theta
y = r sin theta
and theta = inverse tan (x/y)

How do i get these limits? Thanks.

2. $z = x^2 + y^ 2$ is the error

3. The surface area is

$S=\displaystyle\iint_{x^2+y^2\leq 4}\sqrt{1+4x^2+4y^2}\;dxdy=$

$\displaystyle\int_{0}^{2\pi}d\theta\displaystyle\i nt_{0}^{2}\sqrt{1+4r^2}\;r\;dr=\ldots$

4. Originally Posted by FernandoRevilla
The surface area is

$S=\displaystyle\iint_{x^2+y^2\leq 4}\sqrt{1+4x^2+4y^2}\;dxdy=$

$\displaystyle\int_{0}^{2\pi}d\theta\displaystyle\i nt_{0}^{2}\sqrt{1+4r^2}\;r\;dr=\ldots$
According to my book its 3 and 0.

How did you calculate the limits, i don't understand?

According to my book its 3 and 0.

What is $3$ and $0$?. Could you rewrite the question?. Your first post says the plane $z=4$, so the projection over the $xy$ plane is $x^2+y^4\leq 4$ .

6. Originally Posted by FernandoRevilla
What is $3$ and $0$?. Could you rewrite the question?. Your first post says the plane $z=4$, so the projection over the $xy$ plane is $x^2+y^4\leq 4$ .
Your 2nd integral you have got between 2 and 0 my book says its between 3 and 0

7. Can you just tell me how to work it out please? Thanks.

Your 2nd integral you have got between 2 and 0 my book says its between 3 and 0

Can you just tell me how to work it out please? Thanks.
Better fifty fifty.

First question:

Could you please write the projection of the given surface over the $xy$ plane ?

10. Originally Posted by FernandoRevilla
Better fifty fifty.

First question:

Could you please write the projection of the given surface over the $xy$ plane ?
It just says

Find the surface area of the circular paraboloid z = x^2 + y^2 between the xy-plane and the plane z= 4