Question

Surface/area integrals. Evaluate the double integral zdS over the area of a cone, z=sqrt(x^2+y^2) between z=0 and z=1. So, the area I have to calculate is that of a circle with r=1. I transformed the variables to the spherical coordinates with x=cos theta and y = sin theta. Using the limits for r:[0,1] and theta:[0,2pi]... start of the area integral in next post

Answer 1

\[\int\limits_{0}^{2 \pi}\int\limits_{0}^{1}r ^{2}\sqrt{1+?+?} dr d \theta\]

Not sure what I need to put for the derivatives over r and theta

Answer 2

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx

Answer 3

So far I have

\[\int\limits_{}^{}\int\limits_{s}^{}\sqrt{x^2+y^2}.\sqrt{1+4(x^2+y^2)} dxdy\]

Then I plug in the transformation variables instead and their limits

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{1} r^2 \sqrt{1+4r^2} dr d \theta\]

Now I have something I can try to solve.

Answer 4

z = f(x,y) = sqrt(x^2+y^2)

f_x = x/sqrt(x^2+y^2)
f_y = y/sqrt(x^2+y^2)

(f_x)^2 + (f_y)^2 = 1

Answer 5

right ?

Answer 6

\[\large S = \iint\limits_D \sqrt{f_x^2 + f_y^2 +1} dxdt \]

Answer 7

yes

Answer 8

No, not just the root alone

Answer 9

\[\large S = \iint\limits_D \sqrt{1 +1} dxdy \]

Answer 10

\[\large S =\sqrt{2} \iint\limits_D 1 dxdy \]

Answer 11

the root stands for the vector product. It's a normal.

Answer 12

you still have to multiply it with the function itself

Answer 13

hmm work it in rectangular coordinates, its easy

Answer 14

Hmm, I think I'm mixing it up with something else

Answer 15

This is the surface area of a surface `z = f(x,y)` formula :

\[\large S = \iint\limits_D \sqrt{f_x^2 + f_y^2 +1} dxdy\]

Answer 16

I have an example where z=1+2x+3y and the double integral is

\[\int\limits_{}^{}\int\limits_{S}^{}x ^{2}yz dS\]

Substitute the expression for z in the integral function and then you multiply with the squareroot of the partial derivatives you get out of the z expression.

For the example in this post that would give

\[\int\limits_{0}^{3}\int\limits_{0}^{2}x ^{2}y(1+2x+3y)\sqrt{2^{2}+3^{2}+1} dydx\]

Answer 17

It's the function x squareroot of the partial derivatives. You use the squareroot not to replace z, but to get rid of the dz

Answer 18

My source for the example I gave is from a youtube lecture by Dr Cris Tisdell

Answer 19

My course

\[\int\limits_{}^{}\int\limits_{\sum}^{} \left( F,d \sigma \right)=\int\limits_{}^{}\int\limits_{D}^{}\left( F(x(u,v)),x_u X x_v \right) dudv\]

Answer 20

it's an inproduct

Answer 21

I was wrong... that version again is for a vectorfield.
You were correct. I just have to do the squareroot.

I had this for the root alone

\[\sqrt{1+\frac{ x² }{ x^2+y^2 }+\frac{ y^2}{ x^2+y^2 }}\]

I replace it all with the spherical variables

\[\sqrt{1+\frac{ r^2\cos^2\theta }{r^2}+\frac{ r^2\sin^2\theta }{r^2}} \]

The r^2 eliminate each other, and then you have cos^2 + sin ^2 = 1
so the root becomes

\[\sqrt{2}\]

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{1}r \sqrt{2} drd \theta = \sqrt{2}\]

Answer 22

sorry the solution is

\[\sqrt{2} \pi\]

Answer 23

nice :) we can do this using geometry also :

\[\large \begin{align} S &= \iint\limits_D \sqrt{1+f_x^2 + f_y^2 } dxdy\\~\\

&=\iint\limits_D \sqrt{1+\frac{ x^2}{ x^2+y^2 }+\frac{ y^2}{ x^2+y^2 }} dxdy\\~\\
&=\iint\limits_D \sqrt{1+\frac{ x^2+y^2 }{ x^2+y^2 }} dxdy\\~\\
&=\iint\limits_D \sqrt{1+1} dxdy\\~\\
&= \sqrt{2} \iint\limits_D1dxdy\\~\\
&= \sqrt{2} \pi~~~\color{gray}{(\because \text{Area of unit circle = }\pi*1^2 = \pi)}\\~\\

\end{align}\]