How to find volume under a surface using a double integral

To find the volume under a surface, we can add up its height over the region below it. We’ll sketch that region first so the picture tells us where each integral starts and ends.

Write the height above the base

If the surface z=f(x,y) is nonnegative over a region D and the bottom is the xy-plane, the volume is the double integral of f over D. If the solid sits between two surfaces, use the upper height minus the lower height wherever the upper one stays above the lower one.

V=∬Df(x,y)dA

Sketch the triangle and read the bounds

Find the volume under z=2xy above the triangle bounded by x=1, y=1, and y=3-x. The lines meet at (1,1), (2,1), and (1,2). For example, substituting y=1 into y=3-x gives x=2.

The triangular base for the volume integralThe triangle has vertices one comma one, two comma one, and one comma two. A purple vertical slice starts on y equals one and ends on y equals three minus x.(1, 1)(2, 1)(1, 2)Dy = 3 − xy = 1x = 1xy
For a fixed x, the purple slice starts at y = 1 and ends at y = 3 − x.

With vertical slices, x runs from 1 to 2. At each fixed x, y runs from 1 up to 3-x. Integrate with respect to y first.

V=∫12∫13-x2xydydx

Evaluate the inner integral and then the outer integral

Hold x fixed. The antiderivative of 2xy with respect to y is xy2. Evaluate it at 3-x and 1, then simplify the expression in x.

V=∫12(x(3-x)2-x)dx=∫12(8x-6x2+x3)dx

Now integrate with respect to x, and subtract the value at 1 from the value at 2.

V=[4x2-2x3+14x4]12=4-94=74

The volume is 7/4 cubic units.

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