# Polar Co-ordinates

• Apr 13th 2011, 08:21 AM
Polar Co-ordinates
Find the surface area of the circular paraboloid
$\displaystyle z = x^2+ y^2$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.
• Apr 13th 2011, 08:25 AM
$\displaystyle z = x^2 + y^ 2$ is the error
• Apr 13th 2011, 08:35 AM
FernandoRevilla
The surface area is

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

$\displaystyle \displaystyle\int_{0}^{2\pi}d\theta\displaystyle\i nt_{0}^{2}\sqrt{1+4r^2}\;r\;dr=\ldots$
• Apr 13th 2011, 08:43 AM
Quote:

Originally Posted by FernandoRevilla
The surface area is

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

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

According to my book its 3 and 0.

How did you calculate the limits, i don't understand?
• Apr 13th 2011, 08:58 AM
FernandoRevilla
Quote:

Originally Posted by adam_leeds
According to my book its 3 and 0.

What is $\displaystyle 3$ and $\displaystyle 0$?. Could you rewrite the question?. Your first post says the plane $\displaystyle z=4$, so the projection over the $\displaystyle xy$ plane is $\displaystyle x^2+y^4\leq 4$ .
• Apr 13th 2011, 09:10 AM
Quote:

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

Your 2nd integral you have got between 2 and 0 my book says its between 3 and 0
• Apr 13th 2011, 09:26 AM
Can you just tell me how to work it out please? Thanks.
• Apr 13th 2011, 10:01 AM
FernandoRevilla
Quote:

Originally Posted by adam_leeds
Your 2nd integral you have got between 2 and 0 my book says its between 3 and 0

Your book is wrong.
• Apr 13th 2011, 10:03 AM
FernandoRevilla
Quote:

Originally Posted by adam_leeds
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 $\displaystyle xy$ plane ?
• Apr 13th 2011, 10:19 AM
Quote:

Originally Posted by FernandoRevilla
Better fifty fifty.

First question:

Could you please write the projection of the given surface over the $\displaystyle xy$ 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
• Apr 13th 2011, 10:33 AM
FernandoRevilla
Quote:

Originally Posted by adam_leeds
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

Well, look at the answer #3 and tell me exactly what you don't undesrtand.