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 \(R\). \(f(x, y)=\sqrt{(x-y)(x+4 y)}\) \(R:\) region bounded by the parallelogram with vertices \((0,0),\) (1,1),(5,0),(4,-1)

Short answer

To find the volume, change to the variables \(u = x-y\) and \(v = x+y\), compute the Jacobian determinant, rewrite and compute the double integral with respect to these new variables.

Step-by-step solution

Determine the change of variables

After visualizing the given parallelogram, we realize that an appropriate change of variables could be \(u=x - y\) and \(v=x + a y\), where \(a\) represents the slope of the line formed by the points \((0, 0)\) and \((1, 1)\). Since these points lie on a line of slope \(1\), we can use \(a=1\) in our change of variables.

Compute the Jacobian of the transformation

The Jacobian determinant of the transformation (\(u=x - y\), \(v=x + a y\)) is given by the determinant of the matrix of first derivatives, \(J(u, v) =\frac{\partial(x,y)}{\partial(u,v)}= \det \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}\), where \(u(x, y)\) and \(v(x, y)\) have to be inverted. Here, solving \(u=x - y\) and \(v=x + a y\) gives \(x=\frac{v+u}{a+1}\), \(y=\frac{v-u}{a+1}\). The Jacobian determinant will then simplify to \(\frac{1}{a+1}\). For \(a=1\), the Jacobian determinant is \(\frac{1}{2}\).

Rewrite the integral with new variables and compute

The triple integral computing the volume of the solid region becomes \(\int \int \int f(u, v)|J(u, v)| dxdydz = \int \int \int f(v+u, v-u) | \frac{1}{2}| dudvdz\). Also, the limits for \(u\) and \(v\) are \(u:v-1 \leq u \leq v+1\) and \(v: 0 \leq v \leq 5\). We compute this resulting double integral to find the volume of the solid.