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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.

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Solution Summary

Finding the volume of a solid region by using double integration is a standard problem in multi-variable calculus. This solution demonstrates the method by means of an example worked out in detail. A diagram is included showing the region of integration (attached as a PNG image). Additionally, this solution includes references for further reading on the theory behind this question.

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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 - ...

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