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    Volume of Solid by Double Integration

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    Find the volume of the solid in the first octant bounded by the surfaces of z = 1 - y^2, y = 2, and x = 3.

    © BrainMass Inc. brainmass.com December 24, 2021, 9:14 pm ad1c9bdddf
    https://brainmass.com/math/integrals/volume-solid-double-integration-357964

    SOLUTION This solution is FREE courtesy of BrainMass!

    Given a function y = f(x) of a single variable x, we know that the definite integral:

    Integral { f(x) dx, x = a to b }

    gives the area under the curve y = f(x), between the portions x = a and x = b.

    Similarly, for a function z = f(x, y) of two variables x and y, we can use double integration to find the volume under the surface defined by z = f(x, y) over a rectangular portion of the x-y plane (this rectangular region being defined by the limits of integration).

    Here, the function is:

    z = 1 - y^2

    The limits of integration are not directly specified, but it is mentioned that the region is bounded by the above surface and the planes y = 2 and x = 3, and lies in the first octant.

    The first octant is mathematically specified by the three inequalities:

    x >=0
    y >= 0
    z >= 0

    Thus the lower limits of the double integral must be x = 0 and y = 0.

    The upper limits are given by the planes y = 2 and x = 3. However, note that in the first octant, z should also be non-negative. But the value of z is negative for y > 1. (When y = 1, z = 0, and when y > 1, 1 - y^2 < 0)
    Thus, the upper limit for y in the integral must be y = 1.

    This is easily seen if the function is plotted. A plot is shown in the image file attached. The orange-white mesh shows the co-ordinate planes, with the y-axis as the horizontal and the z-axis as the vertical (the x-axis being along the depth).

    Thus the double integral which gives the volume is:

    Integral { 1 - y^2 dy dx, y = 0 to 1, x = 0 to 3} (See image file attached - "Integral.png")

    This integration yields:

    = Integral { [y - (y^3)/3, y = 0 to 1] dx, x = 0 to 3} (as the inner integral is evaluated first)
    = Integral {2/3 dx, x = 0 to 3}
    = [2x/3, x = 0 to 3]
    = 2

    References:
    For some basic theory and examples, refer: http://ltcconline.net/greenl/courses/202/multipleintegration/Volume.htm
    For an overview of the theory of different kinds of multiple integrals, see: http://en.wikipedia.org/wiki/Multiple_integral

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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:14 pm ad1c9bdddf>
    https://brainmass.com/math/integrals/volume-solid-double-integration-357964

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