Question

Use double integrals to find the indicated volumes. Above the rectangle with vertices \((0,0),(0,1),(2,0),\) and \((2,1),\) and below the surface \(z^{2}=36 x^{2}\left(4-x^{2}\right)\).

Short answer

The volume is 16 cubic units.

Step-by-step solution

Understand the region and function

The given surface equation is \(z^{2}=36 x^{2}\big(4-x^{2}\big)\). The region of integration is the rectangle with vertices (0,0), (0,1), (2,0), and (2,1). This implies bounds for \(x\) from 0 to 2 and for \(y\) from 0 to 1.

Solve for \(z\)

Start by solving for \(z\). Since \(z^{2}=36 x^{2}(4-x^{2})\), take the square root of both sides to get \(z = 6x \big(4 - x^{2}\big)^{1/2}\) (considering the positive square root as it represents the surface above the xy-plane).

Set up the double integral

The volume is found by integrating the function describing the surface over the given region. Thus, set up the double integral as follows: \[V = \int_{0}^{1}\int_{0}^{2} 6x \big( 4 - x^{2} \big)^{1/2} \,dxdy\]

Evaluate the inner integral

Integrate with respect to \(x\). Let \(u = 4 - x^2\), then \(du = -2x \, dx\). This transforms the integral as follows: \[\int_{0}^{2} 6x (4-x^{2})^{1/2} \,dx = \int_{4}^{0} -3 (u^{1/2}) \, du = 3 \int_{0}^{4} u^{1/2} \, du\]Integrate with respect to \(u\): \[3 \left[ \frac{2}{3} u^{3/2} \right]_{0}^{4} = 3 \left( \frac{2}{3} \left( 4^{3/2} - 0 \right) \right) = 3 \left( \frac{2}{3} \left( 8 \right) \right) = 16\]

Evaluate the outer integral

Now integrate the result with respect to \(y\):\[V = \int_{0}^{1} 16 \, dy = 16y \bigg|_{0}^{1} = 16 \]

Key concepts

Volume Calculation

When using double integrals to find a volume, you're essentially summing up infinitesimally small pieces of volume under a surface and above a certain region. In this exercise, the region is the rectangle with vertices \(0,0\), \(0,1\), \(2,0\), and \(2,1\), and the surface is given by the equation \(z^{2}=36 x^{2}(4-x^{2})\). The process involves integrating the function that describes the surface over the specified region.
Start by understanding the surface and the region. Here, the surface is a function of \(x\), and the region is a simple rectangle. This makes the volume calculation straightforward because the boundaries are well-defined.

Integration Bounds

Integration bounds define the region over which the integration takes place. For this exercise, the region is a rectangle in the xy-plane with vertices \(0,0\), \(0,1\), \(2,0\), and \(2,1\). This implies the following bounds:

• For \(x\): 0 to 2
• For \(y\): 0 to 1

These bounds help set up our double integral correctly. We first integrate with respect to \(x\) from 0 to 2, and then with respect to \(y\) from 0 to 1. It's crucial to confirm these bounds from the given vertices to avoid errors in the volume calculation.

Variable Substitution

Variable substitution, or change of variables, is a useful technique to simplify the integration process. For the inner integral in this exercise, we use a substitution to make the integral more manageable. Consider the inner integral:
To evaluate \(\[\begin{equation} \int_{0}^{2} 6x(4-x^2)^{1/2} \, dx \end{equation}\]\), let \(u = 4 - x^2\). Then the differential \du = -2x \, dx\. Substituting these into the integral, we get:

• Change of limits: When \(x = 0\), then \(u = 4\); when \(x = 2\), then \(u = 0\).
• Transformed integral: \( \int_{0}^{2} 6x(4-x^2)^{1/2} \, dx = \, - \int_{4}^{0} 3u^{1/2} \, du \)
• Rewriting the limits: \( \int_{4}^{0} -3u^{1/2} \, du = \, 3 \int_{0}^{4} u^{1/2} \, du \)

Now, the integral is simpler to solve.

Evaluating Integrals

Once we have substituted and transformed the integral, it comes down to evaluating it:

• The inner integral becomes \( \int_{4}^{0} -3u^{1/2} \, du = 3 \int_{0}^{4} u^{1/2} \, du = 3[ \, \frac{2}{3}u^{3/2} \, ]_{0}^{4} \)
• Which simplifies to \(3 \,\times \, \frac{2}{3} \,\times \, (8 - 0) = 16\)

The result of the inner integral is 16. Now, we move to the outer integral:
\( \int_{0}^{1} 16 \, dy = 16y \, | \,_{0}^{1}=16\)
Thus, the volume under the surface and above the region is 16 cubic units.