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Math Help - [HELP!] Surface Integrals

  1. #1
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    Exclamation [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.

    Please help~~
    Last edited by violet8804; December 10th 2009 at 08:30 AM. Reason: Question is solved.
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  2. #2
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    Hi. By definition:

    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 g(x,y,z)=z+x^2 y over the surface 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:

    \int_0^3\int_0^1 (\sqrt{1-y^2}+x^2y)\sqrt{f_x^2+f_y^2+1}dydx
    Attached Thumbnails Attached Thumbnails [HELP!] Surface Integrals-cylinder-sa-prob.jpg  
    Last edited by shawsend; December 10th 2009 at 04:09 AM. Reason: corrected formula
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  3. #3
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    Quote Originally Posted by violet8804 View Post
    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.

    Please help~~
    Here's a general concept: if you write the surface as \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}, then the two derivatives, \vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k} and \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), \vec{r}_u\times\vec{r_v}, is perpendicular to the surface and its length gives the "differential of surface area": ||\vec{r}_u\times\vec{r}_v||dudv.

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

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

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

    \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.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Here's a general concept: if you write the surface as \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}, then the two derivatives, \vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k} and \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), \vec{r}_u\times\vec{r_v}, is perpendicular to the surface and its length gives the "differential of surface area": ||\vec{r}_u\times\vec{r}_v||dudv.



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



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



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



    \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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