If R = (x, y) | –1 s x s 1, -5 sy s 5, evaluate the EXAMPLE 2 integral 1- x² dA. It would be very difficult to evaluate this integral directly but, because V1- x? 2 0, we can compute the integral by interpreting it as SOLUTION a volume. If z = V1- x², then x2 + z? = so the given double integral represents the volume of the solid S that lies below the circular cylinder x2 + z² = and z 2 0, and above the rectangle R. (See the figure.) The volume of S is the area of a semicircle with radius times the length of the cylinder. Thus /| V1- x² dA · 10

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