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Math Help - Double integrals question.

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    Double integrals question.

    Find the volume of the solid in the first octant bounded by the coordinates planes, the plane x=3 and the parabolic cylinder z = 4-(y^2).

    Can you please explain how to find limits of x and y in this question as I am not clear how to find them.
    Help would be appreciated.
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    Re: Double integrals question.

    Have you tried drawing a sketch of the region? Also, surely from the given information you can at least see the lower bound for each variable...
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    Re: Double integrals question.

    http://img196.imageshack.us/img196/6752/20130530162.jpg

    here is what I sketched. I think the lower limit of x will be 0 and upper limit will be 3 but I don't know about y
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    Re: Double integrals question.

    Bills
    Your figure is wrong. you are in 3D and the x=3 is not A LINE BUT A PLANE!!!
    SEE THE FIGURE BELOW AND TRY TO DO SOME WORK...DO NOT FORGET THAT YOU MUST RESTRICT YOURSELF IN THE FIRST OCTANT.......the red is the plane x=3 the blue the surface z=...

    Double integrals question.-double-figure.png
    Last edited by MINOANMAN; May 29th 2013 at 11:28 PM.
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    Re: Double integrals question.

    Its to confusing for me I am having difficult time solving these questions.
    here is what I sketched.

    http://img818.imageshack.us/img818/2...0530211915.jpg
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    Re: Double integrals question.

    I have got a little bit now I think I have sketched it again.

    http://img716.imageshack.us/img716/888/paraxa.jpg

    the lower limit of y will be 0 but what about the upper limit?
    Last edited by bllnsr; May 30th 2013 at 10:41 AM.
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    Re: Double integrals question.

    Your graph is not very clear. I take it that parabola is supposed to be the cylinder " z= 4- y^2" but it is drawn in the wrong direction. It goes up to z= 4 , at y= 0 and crosses the y-axis, z= 0, at y= 2. And the cylindrical surface extends parallel to the x-axis.
    It should be clear that x goes from 0 to 4, that for each x, y goes from 0 to 2, and that, for each x and y, z goes from 0 to z= 4- y^2.

    Double integrals question.-volume.jpg
    Last edited by HallsofIvy; May 30th 2013 at 11:22 AM.
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    Re: Double integrals question.

    wont x go from 0 to 3?
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    Re: Double integrals question.

    Find first the area of the parabolic sector as int_0^2 (4-y^2)dy, then extrude it from x=0 to x=3. so V=(int_0^3(int_0^2 (4-y^2)dy)dx).
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    Re: Double integrals question.

    Quote Originally Posted by bllnsr View Post
    wont x go from 0 to 3?
    Yes, I wrote that wrong.
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    Re: Double integrals question.

    The "footprint" in the xy- plane is a rectangle: x goes from 0 to 3 and y goes from 0 to 2.
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    Re: Double integrals question.

    Quote Originally Posted by mopen80 View Post
    Find first the area of the parabolic sector as int_0^2 (4-y^2)dy, then extrude it from x=0 to x=3. so V=(int_0^3(int_0^2 (4-y^2)dy)dx).
    Of course, this only works because the solid formed is a prism, in other cases you will need to perform a double integral.
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    Re: Double integrals question.

    Quote Originally Posted by bllnsr View Post
    Find the volume of the solid in the first octant bounded by the coordinates planes, the plane x=3 and the parabolic cylinder z = 4-(y^2).

    Can you please explain how to find limits of x and y in this question as I am not clear how to find them.
    Help would be appreciated.
    Here's the full graph.
    Graph from the side:

    Double integrals question.-triple_integral3.jpg

    Graph tilted to the right:
    Double integrals question.-triple_integral.jpg

    same graph tilted to the left:
    Double integrals question.-triple_integral2.jpg


    Here z = 4 - y^2 is cyan
    x = 3 is dark blue
    x=0 is red
    y = 0 is gray
    and z = 0 is yellow.

    You need to find the volume of the area that is enclosed with cyan, dark blue, red, gray and yellow parts of the graph which is in the first octant.
    Last edited by x3bnm; June 2nd 2013 at 01:06 PM.
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