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  1. #1
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    volume of a solid

    find the volume of a solid in the first octant bounded above the plane z=5-2y and below by the rectangle in the xy-plane {(x,y):0<= x<=3,0<=y<=2}

    i am lost about setting up the integral can any one help me???
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chris25 View Post
    find the volume of a solid in the first octant bounded above the plane z=5-2y and below by the rectangle in the xy-plane {(x,y):0<= x<=3,0<=y<=2}

    i am lost about setting up the integral can any one help me???
    Volume of a solid would be \int_a^b\int_c^d f(x,y) \,dy\,dx

    In this case, f(x,y)=z=5-2y.

    You're given restrictions on the region R: \left\{R: \ 0\leq x \leq3, \ 0\leq y \leq 2\right\}. This defines the x and y limits of integration.

    Do you think you can set up the volume integral now?

    --Chris
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Volume of a solid would be \int_a^b\int_c^d f(x,y) \,dy\,dx

    In this case, f(x,y)=z=5-2y.

    You're given restrictions on the region R: \left\{R: \ 0\leq x \leq3, \ 0\leq y \leq 2\right\}. This defines the x and y limits of integration.

    Do you think you can set up the volume integral now?

    --Chris
    Generalising:
    When you have a volume bounded above by z = f(x,y) and below by z = g(x,y), then V = \int \int_{R_{xy}} f(x,y) - g(x,y) \, dx \, dy where R_{xy} is the region in the xy-plane that the volume projects onto.

    Note: In this question, f(x,y) = 5 - 2y and g(x,y) = 0 so the volume is given by V = \int \int_{R_{xy}} 5 - 2y \, dx \, dy, the result given by Chris L T521.
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  4. #4
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    i set it up and when i evaluated the integral i received a volume of 18...is this correct?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chris25 View Post
    i set it up and when i evaluated the integral i received a volume of 18...is this correct?


    That is correct.

    --Chris
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