# Thread: [HELP!] Surface Integrals

1. ## [SOLVED] Surface Integrals

I don't know how to use the MATH code here...and I'm running out of time for reviewing for finals, so forgive me if it isn't clear for you... >.<

OK...here's my question:

DoubleInt(s) (z+x^2y)dS, S is the part of the cylinder y^2 + z^2 = 1 that lies between the planes x = 0 and x = 3 in the first octant.

I know that maybe...I should use
y = rcos(theta) = cos(theta)
z = rsin(theta) = sin(theta)
x = x
and find r_(theta) x r_(z) = < cos(theta), sin(theta), o> (I'm not sure if this way is the right path...)
and find the magnitude of |r_(theta) x r_(z)| = 1

I don't know what to do next, which formula to use (there were two...) and what boundaries? How can I get the numbers for those boundaries?

Can anyone solve this question for me please? I know the final answer, which is 12.

2. Hi. By definition:

$\displaystyle S=\mathop\int\int\limits_{\hspace{-15pt}S} g(x,y,z)dS=\mathop\int\int\limits_{\hspace{-15pt}R} g(x,y,f(x,y))\sqrt{f_x^2+f_y^2+1} dA$

So you're integrating the function $\displaystyle g(x,y,z)=z+x^2 y$ over the surface $\displaystyle f(x,y)=z=\sqrt{1-y^2}$ between the planes x=0 and x=3 in the first quadrant. That's just the top one-quarter of the cylinder along the x-axis shown in red below. Now, the region in the x-y plane below that part, since the cylinder has a radius of one, would be 0 to 3 in the x-direction and 0 to 1 in the y-direction. Now, turn the crank:

$\displaystyle \int_0^3\int_0^1 (\sqrt{1-y^2}+x^2y)\sqrt{f_x^2+f_y^2+1}dydx$

3. Originally Posted by violet8804
I don't know how to use the MATH code here...and I'm running out of time for reviewing for finals, so forgive me if it isn't clear for you... >.<

OK...here's my question:

DoubleInt(s) (z+x^2y)dS, S is the part of the cylinder y^2 + z^2 = 1 that lies between the planes x = 0 and x = 3 in the first octant.

I know that maybe...I should use
y = rcos(theta) = cos(theta)
z = rsin(theta) = sin(theta)
x = x
and find r_(theta) x r_(z) = < cos(theta), sin(theta), o> (I'm not sure if this way is the right path...)
and find the magnitude of |r_(theta) x r_(z)| = 1

I don't know what to do next, which formula to use (there were two...) and what boundaries? How can I get the numbers for those boundaries?

Can anyone solve this question for me please? I know the final answer, which is 12.

Here's a general concept: if you write the surface as $\displaystyle \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$, then the two derivatives, $\displaystyle \vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\displaystyle \vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ are tangent to the surface and their cross product (called the "fundamental vector product" of the surface), $\displaystyle \vec{r}_u\times\vec{r_v}$, is perpendicular to the surface and its length gives the "differential of surface area": $\displaystyle ||\vec{r}_u\times\vec{r}_v||dudv$.

Yes, you can use $\displaystyle y= cos(\theta)$, $\displaystyle z= sin(\theta)$, x= x with $\displaystyle \theta$ and x as parameters u and v. That is, $\displaystyle \vec{r}= x\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$ so $\displaystyle \vec{r}_x= \vec{i}$ and $\displaystyle \vec{r}_\theta= -sin(\theta)\vec{j}+ cos(\theta)\vec{k}$ so that the "fundamental vector product" is

$\displaystyle \vec{r}_x\times\vec{r}_\theta= -cos(\theta)\vec{i}- sin(\theta)\vec{k}$.

Its length is, of course, 1 so $\displaystyle dV= dxd\theta$. The function to be integrated is $\displaystyle z+ x^2y= sin(\theta)+ x^2cos(\theta)$. In the first octant, both y and z are positive so $\displaystyle \theta$ goes from 0 to $\displaystyle \pi/2$ and, of course, x goes from 0 to 1. The integral is

$\displaystyle \int_{x=0}^1\int_{\theta= 0}^{\pi/2} sin(\theta)+ x^2 cos(\theta) dxd\theta$.

By the way, if you click on a mathematical expression in any post, you can see the LaTex code used.

4. Originally Posted by HallsofIvy
Here's a general concept: if you write the surface as $\displaystyle \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$, then the two derivatives, $\displaystyle \vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\displaystyle \vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ are tangent to the surface and their cross product (called the "fundamental vector product" of the surface), $\displaystyle \vec{r}_u\times\vec{r_v}$, is perpendicular to the surface and its length gives the "differential of surface area": $\displaystyle ||\vec{r}_u\times\vec{r}_v||dudv$.

Yes, you can use $\displaystyle y= cos(\theta)$, $\displaystyle z= sin(\theta)$, x= x with $\displaystyle \theta$ and x as parameters u and v. That is, $\displaystyle \vec{r}= x\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$ so $\displaystyle \vec{r}_x= \vec{i}$ and $\displaystyle \vec{r}_\theta= -sin(\theta)\vec{j}+ cos(\theta)\vec{k}$ so that the "fundamental vector product" is

$\displaystyle \vec{r}_x\times\vec{r}_\theta= -cos(\theta)\vec{i}- sin(\theta)\vec{k}$.

Its length is, of course, 1 so $\displaystyle dV= dxd\theta$. The function to be integrated is $\displaystyle z+ x^2y= sin(\theta)+ x^2cos(\theta)$. In the first octant, both y and z are positive so $\displaystyle \theta$ goes from 0 to $\displaystyle \pi/2$ and, of course, x goes from 0 to 1. The integral is

$\displaystyle \int_{x=0}^1\int_{\theta= 0}^{\pi/2} sin(\theta)+ x^2 cos(\theta) dxd\theta$.

By the way, if you click on a mathematical expression in any post, you can see the LaTex code used.

Thank you!
And, by the way...
Is the (theta) always from 0 to 2*Pi if it's in the 1st octant (after change to polar coordinates)?
Also, can you please teach me when to use which of the two formula, is there a trick to see from the question when to use which formula?
I always get confused which one to use, they are all the homework questions, and i got partial solutions (the library only have the older version of the solution manual...); so, when I do the question myself, and checked from the solution manual, it used a different formula (or different method) to solve the surface integral...I'm so confused T_T

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### evaluate the surface integral s is part of the cylinder

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