Results 1 to 4 of 4

Math Help - Surface Integral

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    107

    Surface Integral

    Hey I'm stuck on this problem. I don't know if I'm doing something wrong with the surface integral or not because after simplification, I got to an integral that I can't do. Here is the question:

    Evaluate the surface integral:

    \int\int_S yz dS
    S is the surface with parametric equations x = u^2, y = usinv, z = ucosv 0 <= u <= 1, 0 <= v <= \pi/2

    This is attempt:

    \vec{r}_u = <2u, sinv, cosv>

    \vec{r}_v = <0, ucosv, -usinv>

    \vec{r}_u \times \vec{r}_v = <-u, 2u^2sinv, 2u^2cosv>

    |\vec{r}_u \times \vec{r}_v| = u\sqrt{4u^2 + 1}

    \int \int_D (usinv)(ucosv)|\vec{r}_u \times \vec{r}_v| dA

    \int_0^1 u^3\sqrt{4u^2+1}du \int_0^{\pi/2} sinv cosvdv

    I don't know how to evaluate the integral \int_0^1 u^3\sqrt{4u^2+1}du :\. if my steps are incorrect, I would be obliged if someone can tell me what I'm doing wrong and how to evaluate that integral.

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by lilaziz1 View Post
    Hey I'm stuck on this problem. I don't know if I'm doing something wrong with the surface integral or not because after simplification, I got to an integral that I can't do. Here is the question:

    Evaluate the surface integral:

    \int\int_S yz dS
    S is the surface with parametric equations x = u^2, y = usinv, z = ucosv 0 <= u <= 1, 0 <= v <= \pi/2

    This is attempt:

    \vec{r}_u = <2u, sinv, cosv>

    \vec{r}_v = <0, ucosv, -usinv>

    \vec{r}_u \times \vec{r}_v = <-u, 2u^2sinv, 2u^2cosv>

    |\vec{r}_u \times \vec{r}_v| = u\sqrt{4u^2 + 1}

    \int \int_D (usinv)(ucosv)|\vec{r}_u \times \vec{r}_v| dA

    \int_0^1 u^3\sqrt{4u^2+1}du \int_0^{\pi/2} sinv cosvdv

    I don't know how to evaluate the integral \int_0^1 u^3\sqrt{4u^2+1}du :\. if my steps are incorrect, I would be obliged if someone can tell me what I'm doing wrong and how to evaluate that integral.

    Thanks in advance!
    let \displaystyle t=4u^2+1 \iff u^2=\frac{t-1}{4} \implies 2udu=\frac{dt}{4}

    This gives

    \displaystyle \int_{1}^{5}\left(\frac{t-1}{4} \right)\sqrt{t}\left( \frac{dt}{8}\right)=\frac{1}{32}\int_{1}^{5}t^{\fr  ac{3}{2}}-t^{\frac{1}{2}}dt
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2010
    Posts
    107
    ooh alright. I got another question. How do you look at an integral like that and be like "I have to do substitution here!" If I had an integral like \int x\sqrt{4x^2 +1}dx, I can see that I have to do substitution, but for the integral in the first post, I wouldn't know where to start.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,417
    Thanks
    1329
    Quote Originally Posted by lilaziz1 View Post
    ooh alright. I got another question. How do you look at an integral like that and be like "I have to do substitution here!" If I had an integral like \int x\sqrt{4x^2 +1}dx, I can see that I have to do substitution, but for the integral in the first post, I wouldn't know where to start.
    What TheEmptySet did was see that " 4x^2+ 1" in the square root and look for an "x" to put with the dx!

    Instead of x outside the square root there is x^3 but TheEmptySet saw that that was x^2(x) so he took out one x for the "x dx".
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 5th 2011, 05:43 AM
  2. Surface Integral
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 19th 2011, 05:50 AM
  3. Surface Integral
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 3rd 2011, 04:28 AM
  4. Volume integral to a surface integral ... divergence theorem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: October 7th 2010, 06:11 PM
  5. Surface Integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 31st 2008, 02:26 AM

Search Tags


/mathhelpforum @mathhelpforum