Volume of Solid by Double Integration

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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