1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface Area Multiple Integrals problem

  1. Mar 24, 2007 #1
    Hi,
    I need some help on these problems. I'm not sure what to do.

    1 Find the area of the plane with vector equation r(u, v) =< 1+v, u-2v, 3-5u+v> that is given by 0<u<1, 0<v<1.

    So far, I took the partial derivatives with respect to u and v. I don't know if I was supposed to or not and I'm really stuck there.

    Since I'm trying to find surface area a relevant equation may be S = (double integral) SQRT(1 + (partial derivative of u)^2 + (partial derivative of v)^2) dA

    2. The part of the surface z = 1 + 3x +2y^2 that lies above the triangle with vertices (0,0), (0,1), (2,1)

    Relevant equation should be same as above.

    I think I got the answer but I'm not sure and would really appreciate some help if I got it wrong.
    Here's how I did it:
    partial x: 3 Partial y: 4y
    The boundary of the triangle is x=2y
    so it took the integral from 0 to 1 and from 0 to 2y SQRT(10 + 16y^2)dx dy
    and I got the integral from 0 to 1 times 2y SQRT(10 +16y^2) dy
    then I took that integral and got (2/3)*(2/32)*SQRT(10 + 16Y^2)^(3/2) from 0 to 1
    My final answer is (1/24)(26)^(3/2) - (1/24)(10)^(3/2)

    Is this correct?
     
  2. jcsd
  3. Mar 24, 2007 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Remember that when you parametrize a surface, tangent vectors are gained by differentiating with respect to the two independent variables.
    Let us take the general case.

    Suppose you have a surface S parametrized as
    [tex]\vec{S}(u,v)=(X(u,v),Y(u,v),Z(u,v))[/tex]
    Then, two infinitesemal tangent vectors to this surface are given by:
    [tex]d\vec{t}_{u}=\frac{\partial\vec{S}}{\partial{u}}du,d\vec{t}_{v}=\frac{\partial\vec{S}}{\partial{v}}dv[/tex]
    The differential area spanned by these two vectors lies on S, and its area is given by:
    [tex]dA=||d\vec{n}_{S}||=||d\vec{t}_{u}\times{d}\vec{t}_{v}||=||\frac{\partial\vec{S}}{\partial{u}}\times\frac{\partial\vec{S}}{\partial{v}}||dudv[/tex]
    And the total area of S is then gained by summing together all dA's (i.e, by integration).

    Thus, yes, you have thought correctly on 1.
     
  4. Mar 24, 2007 #3
    The answer I got for number 1 is SQRT (107), is this correct?

    What about number 2?
     
  5. Mar 24, 2007 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Hmm..I have to retract what I said about you having thought correctly on 1.

    What you have written is totally meaningless!

    Follow what I wrote in my post:
    [tex]\frac{\partial\vec{r}}{\partial{u}}=\vec{j}-5\vec{k},\frac{\partial\vec{r}}{\partial{v}}=\vec{i}-2\vec{j}+\vec{k}[/tex]
    Thus, the norm of the cross product is:
    [tex]||\frac{\partial\vec{r}}{\partial{u}}\times\frac{\partial\vec{r}}{\partial{v}}||=\sqrt{62}[/tex]
    and since both u and v varies between 0 and 1, [itex]\sqrt{62}[/itex] is the correct answer.
     
  6. Mar 24, 2007 #5
    I still get SQRT (107). Are you sure about SQRT(62)?
     
  7. Mar 24, 2007 #6

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Hmm..perhaps not. Made a too fast calculation. Sorry about that. square root of 107 it is.
     
  8. Mar 25, 2007 #7
    Thanks! What about number 2?
     
  9. Mar 26, 2007 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ???
    [tex]\left|\begin{array}{ccc}i & j & k \\ 0 & 1 & -5 \\ 1 & -2 & 1\end{array}\right|= i\left|\begin{array}{cc}1 & -5 \\ -2 & 1\end{array}\right|- j\left|\begin{array}{cc}0 & -5 \\1 & 1\end{array}\right|+ k\left|\begin{array}{cc} 0 & 1 \\ 1 & -2\end{array}\right|[/tex]
    = -9 i- 5j - k
    which has length [itex]\sqrt{81+25+1}= \sqrt{107}[/itex]
     
  10. Mar 26, 2007 #9

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Indeed, it is.
    However, as I made the mental calculation (-5)*(-2)=-5, so that the i'th component became -6 rather than -9, I got 62 as my radicand rather than 107.
     
    Last edited: Mar 26, 2007
  11. Mar 26, 2007 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You are starting to take after me. That is NOT a good sign!
     
  12. Mar 26, 2007 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the answer to (1) is [itex]\sqrt{107}[/itex]

    For (2): The part of the surface z = 1 + 3x +2y^2 that lies above the triangle with vertices (0,0), (0,1), (2,1)

    Pretty much the same thing. You can write this as x= u, y= v, z= 1+ 3u+ 2v2, basically using x and y as parameters, so that r= ui+ vj+ (1+ 3u+ 2v2)k. Find ru, rv[/sup] and take their cross product. Integrate over the triangle (which has sides x= 0, y= 1, and y= (1/2)x).
     
  13. Mar 26, 2007 #12

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Another way for 1:

    The surface you're looking for is a parallelogram with vertices at r(0,0), r(1,0), r(0,1) and r(1,1). So two of its non-parallel sides are given by r(1,0)-r(0,0) and r(0,1)-r(0,0). Find the norm of the cross product of these two vectors:

    || [r(1,0)-r(0,0)] x [r(0,1)-r(0,0)] ||

    and this will be the desired area.
     
  14. Mar 27, 2007 #13
    Thanks for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?