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Compute volume below z=x^2+y and above [0,1]x[1,2]

  1. Jul 24, 2014 #1
    1. The problem statement, all variables and given/known data

    compute volume below z=x^2+y and above the rectangle R= [0,1]x[1,2]

    3. The attempt at a solution

    [tex]\int_{0}^{1}\int_{1}^{2} x^2+y dydx[/tex]

    [tex]\int_{0}^{1} [x^2y+\frac{y^2}{2}]\Bigg|_{1}^{2} dx[/tex]

    [tex]\int_{0}^{1} x^2+ \frac{3}{2}dx[/tex]

    [tex]\frac{x^3}{3} + \frac{3x}{2} \Bigg|_{0}^{1} = \frac{11}{6}[/tex]

    checking to make sure I understood the question correctly
  2. jcsd
  3. Jul 24, 2014 #2


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    Homework Helper

    It is all rigth, but do not forget the parentheses next time.

    [tex]\int_{0}^{1}\int_{1}^{2}{( x^2+y )dydx}[/tex]

    [tex]\int_{0}^{1}{ (x^2+ \frac{3}{2})dx}[/tex]

  4. Jul 24, 2014 #3


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    Science Advisor

    Since x and y happen to be "separated" in this integral, you could also do it as
    [tex]\int_0^1\int_1^2 (x^2+ y) dy dx= \int_0^1\int_1^2 x^2 dydx+ \int_0^1\int_1^2 y dy dx[/tex]
    [tex]= \left(\int_0^1 x^2 dx\right)\left(\int_1^2 dy\right)+ \left(\int_0^1 dx\right)\left(\int_1^2 y dy\right)= \left[\frac{1}{3}x^2\right]_0^1\left[y\right]_1^2+ \left[x\right]_0^1\left[\frac{1}{2}y^2\right]_1^2[/tex][tex]= \left[\frac{1}{3}\right]\left[1\right]+ \left[1\right]\left[\frac{3}{2}\right]= \frac{11}{6}[/tex]
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