Problem 20 Use double integrals to find the... [FREE SOLUTION]

Chapter 5: Problem 20

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

Expert verified

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 \(y\):\[V = \int_{0}^{1} 16 \, dy = 16y \bigg|_{0}^{1} = 16 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

91影视