Results 1 to 4 of 4

Math Help - Surface Integrals Implicit

  1. #1
    Senior Member
    Joined
    Oct 2008
    Posts
    393

    Surface Integrals Implicit

    I have to evaluate the given surface integrals \iint_ s 3{y}^{2 }z d\sigma where S is the portion of the cylinder {x}^{2 } +{y}^{2 } =1 between the planes z = 0 and z = 1, and on the positive x side of the yz-plane

    First of all i think i need to find n

    Which is \frac{2xi + 2yj}{4 } I think.

    Now i'm not sure on my next step.
    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 adam_leeds View Post
    I have to evaluate the given surface integrals \iint_ s 3{y}^{2 }z d\sigma where S is the portion of the cylinder {x}^{2 } +{y}^{2 } =1 between the planes z = 0 and z = 1, and on the positive x side of the yz-plane

    First of all i think i need to find n

    Which is \frac{2xi + 2yj}{4 } I think.

    Now i'm not sure on my next step.
    I would parameterize the surface of the cylindar using cylindrical polar coordinates.

    x=1\cdot \cos(\phi) \quad y=1\cdot =sin(\phi) \quad z=z

    This gives

    \mathbf{r}(\phi,z)=\cos(\phi)\mathbf{i}+\sin(\phi)  \mathbf{j}+z\mathbf{k}

    This gives

    \mathbf{r}_\phi \times \mathbf{r}_z=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin(\phi) & \cos(\phi) & 0 \\ 0 & 0 & 1 \end{vmatrix}= \cos(\phi)\mathbf{i}+\sin(\phi)\mathbf{j}

    Now if we find its magnitude we get

    |\mathbf{r}_\phi \times \mathbf{r}_z|=\sqrt{\cos^2(\phi)+\sin^2(\phi)}=1

    So this gives the integral

    \iint 3y^2zd\sigma=3\int_{0}^{1} \int_{0}^{2\pi}sin^2(\phi)zd\phi dz =3\left(\int_{0}^{1}zdz \right)\left(\int_{0}^{2\pi}\sin^2(\phi)d\phi \right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Oct 2008
    Posts
    393
    Quote Originally Posted by TheEmptySet View Post
    \iint 3y^2zd\sigma=3\int_{0}^{1} \int_{0}^{2\pi}sin^2(\phi)zd\phi dz =3\left(\int_{0}^{1}zdz \right)\left(\int_{0}^{2\pi}\sin^2(\phi)d\phi \right)
    I just integrated this and got 3pi/2 but the answer is 3pi/4. Am i right or have i done poor integration.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by adam_leeds View Post
    I just integrated this and got 3pi/2 but the answer is 3pi/4. Am i right or have i done poor integration.
    I misread the original problem statement

    on the positive x side of the yz-plane
    This will put

     \phi \in \left[-\frac{\pi}{2}, \frac{\pi}{2} \right]
    This changes
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: April 3rd 2011, 04:51 AM
  2. Replies: 1
    Last Post: December 6th 2009, 08:43 PM
  3. Surface Integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 11th 2009, 08:01 PM
  4. integrals and implicit - and fast
    Posted in the Calculus Forum
    Replies: 23
    Last Post: April 25th 2008, 03:55 AM
  5. Surface Integrals
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 22nd 2007, 03:07 PM

Search Tags


/mathhelpforum @mathhelpforum