Question

In Exercises 17-22, use a change of variables to find the volume of the solid region lying below the surface \(z=f(x, y)\) and above the plane region \(R\). \(f(x, y)=(x+y)^{2} \sin ^{2}(x-y)\) \(R:\) region bounded by the square with vertices \((\pi, 0),\) \((3 \pi / 2, \pi / 2),(\pi, \pi),(\pi / 2, \pi / 2)\)

Short answer

The volume of the solid region is the exact value of the computed double integral.

Step-by-step solution

Understanding and visualizing the region R and the function f(x,y)

First, plot the points given for the region R to understand its shape. From these points, we can observe that they form a square. Now, also try to understand and visualize \(f(x, y)\) is increasing or decreasing as x and y increase.

Changing variable to simplify the integration

In this situation, it may not be necessary to change the variables from cartesian coordinates (x,y) to polar coordinates (r, theta). We can take \(u=x+y\) and \(v=x-y\) as our new variables. Then we will find the Jacobian determinant to account for the area dilation/ contraction when changing variables.

Setting up the double integral

Now, using the limits for u and v derived from the region R, and the Jacobian, set up the double integral. Plug in the function \(f(u, v)\) into the double integral. Remember to multiply \(f(u,v)\) by the absolute of the Jacobian.

Evaluating the integral

Finally, evaluate the outer and inner integral. Start by integrating the inner function with respect to v and then the result with respect to u. The result will give us the volume of the solid.