Double Integral of x*y^3 + 1 over Surface r=1, tetha 0-Pi, z 0-2: Solving Guide

[kasse]

Int Int (x*y^3 + 1) dS

where S is the surface r=1, tetha from 0 to Pi and z from 0 to 2.

How can I solve this integral? I haven't got a clue.

 

[vsage]

What a peculiar question that asks you to evaluate an integrand in cartesian coordinates but defines the surface in cylindrical. I would start by converting x*y^3 to cylindrical coordinates. You should be able to see the solution more clearly then.

 

[HallsofIvy]

The given surface is half a cylinder. Convert to cylindrical coordinates. Do you know how to find the "differential of surface area"?

 

[kasse]

The given surface is half a cylinder. Convert to cylindrical coordinates. Do you know how to find the "differential of surface area"?

No. But I know that x=rcos(t) and y=rsin(t) in cylindrical coords.

 

[HallsofIvy]

Odd, if someone expects you to be able to do a problem like this then surely they expect you to be able to integrate over a surface area! Perhaps you need to review your text.

Since we are given that r=1, we have $x= cos(\theta)$, $y= sin(\theta)$, and z= z. The "position vector" of any point on the surface is $cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ z\vec{j}$.

The derivative with respcect to $\theta$ is $-sin(\theta)\vec{i}+ cos(\theta)\vec{j}$ and the derivative with respect to z is $\vec{k}$. The "fundamental vector product" is the cross product of those two vectors:$cos(\theta)\vec{i}+ sin(\theta)\vec{j}$ and the length of that gives the "differential of surface area". [itex]\sqrt{cos^2(\theta)+ sin^2(\theta)}= 1[itex]so $d\sigma = d\theta dy$.[/itex][/itex]

 

[kasse]

My book simply says that

Int Int (xy^3+1) dS = Int Int dS + 0 = 2*Pi

I don't understand why...


I converted into cylindrical coords:

Int Int (r^5cos(t)sin((t))^3 + r) dr dt

But what are the limits now?

 

[HallsofIvy]

Had a little too much wine with dinner! There should be no "+ 1" in the square root. $d\sigma= d\theta dz$ is the correct differential.

In these coordinates, xy³+ 1 is $cos(\theta)sin^3(\theta)+ 1$ so you want to find
[tex]\int_{\theta= 0}^\pi\int_{z=0}^2 cos(\theta)sin^3(\theta)dzd\theta[/itex][/tex]

 

[kasse]

I understood it now finding some examples in my book. Thank you so much!