Q. 38 - Calculus

Question

For each double integral in Exercises 33–38, (a) sketch the region , (b) use the specified transformation to sketch the transformed region, and (c) use the transformation to evaluate the integral. 

$\int {\int }_{\Omega }\left(\frac{y^3}{x}-xy\right) dA$, where $\Omega$ is the region in the first and second quadrants that is bounded above by the hyperbola $y^2-x^2=12$, bounded below by the hyperbola $y^2-x^2=3$, and bounded on the left and right by the lines y = −2x and y = 2x, respectively. Use the transformation given by $u=\frac{y}{x}$ and $v=y^2-x^2$. 

Step-by-Step Solution

Answer

(a). The sketch of the region is shown below,

(b). The sketch of the transformed region is shown below,

(c). $\int {\int }_{\Omega }\left(\frac{y^3}{x}-xy\right)dA=0$

Part (a): Step 1: Draw the region

The given region $\Omega$ is said to be bounded by,

$y^2-x^2=12, y^2-x^2=3, y=-2x, y=2x$

Plot the given points to form the trapezoid and name the vertices.

In the region $\Omega$, the equations of boundary curves are,

$$\begin{gathered} AB: y=-2x \\ BC: y^2-x^2=12 \\ CD: y=2x \\ DA: y^2-x^2=3 \end{gathered}$$

Part (b): Step 1: Draw the transformed region.

Consider the new set of variables defined as

$$\begin{gathered} u=\frac{y}{x} \\ v=y^2-x^2 \end{gathered}$$

After solving we get that,

$$\begin{gathered} \sqrt{\frac{v}{u^2-1}}=x \\ y=\sqrt{\frac{v{u}^2}{u^2-1}} \end{gathered}$$

Use these equations to determine the equation of each boundary of the region in terms of u and v.

$$\begin{gathered} AB: y=-2x\Rightarrow u=-2 \\ BC: y^2-x^2=12\Rightarrow v=12 \\ CD: y=2x\Rightarrow u=2 \\ DA: y^2-x^2=3\Rightarrow v=3 \end{gathered}$$

Plot these limits on u v- plane.

Part (c): Step 1: Evaluate the double integral.

Set up the double integral.

$$\begin{gathered} \int {\int }_{\Omega }\left(\frac{y^3}{x}-xy\right)dA=\frac{1}{2}{\int }_{u=-2}^{u=2}{\int }_{v=3}^{v=12}\left(\frac{vu}{u^2-1}\right)dvdu \\ \int {\int }_{\Omega }\left(\frac{y^3}{x}-xy\right)dA=\frac{1}{2}{\int }_{-2}^2\left(\frac{u}{u^2-1}{\int }_3^{12}\left(v\right)dv\right)du \\ \int {\int }_{\Omega }\left(\frac{y^3}{x}-xy\right)dA=\frac{1}{2}{\int }_{-2}^2\left(\frac{u}{u^2-1}{\left[\frac{v^2}{2}\right]}_3^{12}\right)du \\ \int {\int }_{\Omega }\left(\frac{y^3}{x}-xy\right)dA=\frac{135}{4}{\int }_{-2}^2\left(\frac{u}{u^2-1}\right)du \left[\frac{u}{u^2-1} is an odd function\right] \\ \int {\int }_{\Omega }\left(\frac{y^3}{x}-xy\right)dA=\frac{135}{4}\times 0 \\ \int {\int }_{\Omega }\left(\frac{y^3}{x}-xy\right)dA=0 \end{gathered}$$