Help me understand a surface integral question?

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To make everything explicit: Knowing now that $[0,3]\times[0,2]$ is a rectangle in the $xy$-plane, we also have our limits of integration. We write $x^2yx=x^2y(1+2x+3y)=x^2y+2x^3y+3x^2y^2$ and integrate over the region $\{(x,y)|0\leq x\leq 3, 0 \leq y \leq 2\}$. Thus our integral is

$\displaystyle\int_{x=0}^3\int_{y=0}^2(x^2y+2x^3y+3x^2y^2)dydx$.

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MS Mathematical Finance graduate from Boston University. Finance, engineer, cybersecurity. Previously quantitative analyst in structured products group at fixed income money manager. Currently Security Customer Advisor, providing dedicated guidance to clients on vulnerability management program execution, from high-level planning to technical details.

Updated on December 10, 2020

Comments

  • ch-pub
    ch-pub almost 3 years

    The question is:

    Evaluate the surface integral: $$ \iint\limits_S \, x^2yz\ \mathrm{d} S $$ Where S is part of the plane z = 1 + 2x + 3y that lies above the rectangle [0,3] X [0,2]

    I literally just don't understand the notation of this "rectangle". How is [0,3] X [0,2] a rectangle? Normally we are given vertices of some sort of shape, or instead just told something like "the part that lies in the first octant". So, I'm having trouble parameterizing the function in terms of u and v.

    Basically, what I've done so far, in terms of parameterizing is:

    u = x

    v = y

    So, z = 1 + 2u + 3v

    $$ \vec{r} \ = u \vec{i} \ + v \vec{j} \ + (1 + 2u + 3v) \vec{k} \ $$

    $$ \vec{r}_u \ = \vec{i} \ + 2 \vec{k} \ $$

    $$ \vec{r}_v \ = \vec{j} \ + 3 \vec{k} \ $$

    So:

    $$ \vec{r}_u \times \vec{r}_v = -2\vec{i} \ - 3\vec{j} + \vec{k} \ $$

    Then I get stuck around this step:

    $$ \iint\limits_S \, u^2v(1 + 2u + 3v)\sqrt{14}\ \mathrm{d} v \, \mathrm{d} u $$

    I need limits of integration.

    Sorry for any formatting mistakes.

    Thank you for the help.

    • Gyu Eun Lee
      Gyu Eun Lee almost 11 years
      [0,3]x[0,2] is the set of all ordered pairs (x,y) such that 0 \leq x \leq 3 and 0 \leq y \leq 2. If you draw this out in the plane it's a rectangle with vertices (0,0), (3,0), (0,2), (3,2).
    • ch-pub
      ch-pub almost 11 years
      Thank you very much. So, the limits of integration would literally just be 0 <= u <= 3 and 0 <= v <= 2. I've just never seen this notation, and didn't want to assume [0,3] was x and [0,2] was y. Thank you!
  • ch-pub
    ch-pub almost 11 years
    Where did the root(14) go in your integral? I'm sorry. I'm not familiar with integrating a surface integral without reparameterizing in terms of vectors first. I'm just learning these things now.
  • Christian Blatter
    Christian Blatter almost 11 years
    The ${\rm d}\omega=|{\bf r}_u\times{\bf r}_v|\ {\rm d}(u,v)=\sqrt{14}\ {\rm d}(u,v)$ went lost here.
  • Tarius
    Tarius almost 8 years
    I just found the official answer to this question in the book - Stewart Calculus, 7E, question 16.7.009. Which stated that the answer is $ = 171 \sqrt{14} $. How is this so, considering that $\sqrt{14}$ is no longer included in your answer?