Question

Use another order of integration to evaluate \(\int_{1}^{4} \int_{z}^{4 z} \int_{0}^{\pi^{2}} \frac{\sin \sqrt{y z}}{x^{3 / 2}} d y d x d z\)

Short answer

# Short Answer The final answer after changing the order of integration and solving the resulting triple integral is \(-\frac{8}{\pi}(1 - 2\pi)\).

Step-by-step solution

1. Change the order of integration

To change the order of integration, we first need to rewrite the limits for each integral in terms of the other two variables. In this case, we will switch to the following order: \(dy dx dz\). To find the new limits, we observe that the limits of integration for \(x\) are from \(y = z\) to \(y = 4z\), and for \(z\) ranges between 1 to 4. Now, we get the new limits: - For dy: \(0 \le y \le \pi^2\) - For dx: \(z \le x \le 4z\) - For dz: \(1 \le z \le 4\) The new integral becomes: \(\int_{1}^{4}\int_{z}^{4z} \int_{0}^{\pi^2} \frac{\sin \sqrt{y z}}{x^{3 / 2}} dy dx dz\)

2. Evaluate the integral with the new order

Now that we have switched the order of integration, we will proceed by evaluating each integral step by step. First, we evaluate the integral with respect to \(y\): \(\int_{0}^{\pi^2} \frac{\sin \sqrt{y z}}{x^{3/2}} dy = \frac{2}{x^{3/2}} \int_{0}^{\pi^2} \sin \sqrt{y z} dy/2\) Now, let \(u = \sqrt{yz}\). Then \(y = \frac{u^2}{z}\) and \(dy = \frac{2u}{z}\, du\). Therefore, the integral becomes: \(\frac{2}{x^{3/2}} \int_{0}^{\pi\sqrt{z}} \sin u \, du = \frac{2}{x^{3/2}} [-\cos u]_{0}^{\pi \sqrt{z}} = \frac{2}{x^{3 / 2}}(1 - \cos(\pi \sqrt{z}))\) Now, the integral becomes: \(\int_{1}^{4} \int_{z}^{4z} \frac{2}{x^{3 / 2}}(1 - \cos(\pi \sqrt{z})) dx dz\) Second, we evaluate the integral with respect to \(x\): \begin{align*} \int_{z}^{4z} \frac{2}{x^{3/2}}(1 - \cos(\pi \sqrt{z})) dx &= (1 - \cos(\pi \sqrt{z}))\left[-\frac{4}{x^{1/2}}\right]_{z}^{4z}\\ &= -4(1 - \cos(\pi \sqrt{z}))\left(\frac{1}{2\sqrt{z}} - \frac{1}{4}\right) \end{align*} The integral becomes: \(\int_{1}^{4} -4(1 - \cos(\pi \sqrt{z}))\left(\frac{1}{2\sqrt{z}} - \frac{1}{4}\right) dz\) Finally, we evaluate the integral with respect to \(z\): \(\int_{1}^{4} -4(1 - \cos(\pi \sqrt{z}))\left(\frac{1}{2\sqrt{z}} - \frac{1}{4}\right) dz = -\frac{8}{\pi}(9 - 16\cos(\pi)\sin(\pi) - 20 + 18\pi\cos(\pi))\) The final answer is: \(-\frac{8}{\pi}(1 - 2\pi)\)

Key concepts

Triple Integrals

When dealing with triple integrals, you're essentially working through three layers of integration, each one touching upon a different variable. In this type of problem, the main idea is to integrate a function over a three-dimensional space. The dimensions typically represented by variables such as \(x\), \(y\), and \(z\), may relate to physical space, but they can also be abstract in some calculus problems.

To get you started, remember:

• The outermost integral usually pertains to the last variable in the sequence (often denoted like \(dz\) for the range of \(z\)).
• The innermost integral deals with the variable that is integrated first (like \(dy\) in our rearranged order).
• In between, the integrals manage the remaining variable, adjusting according to the changes in limits determined by the others.

Triple integrals require careful setting of limits for each variable, maintaining the order specified. It helps paint a complete picture of how the function behaves across a defined 3D region.

Integration Limits

Integration limits establish where each variable starts and stops over the region you're integrating. In this problem, beginning with limits of integration for one order and switching to another reveals another perspective on the problem, often simplifying the evaluation process.

To break it down:

• \(dy\) spans the simple, fixed limits from \(0\) to \(\pi^2\), no changes as \(y\) doesn’t depend on the other variables here.
• \(dx\) adapts its limits from \(z\) to \(4z\), responding directly to the set conditions tied to \(z\). It's these limits that indicate \(x\) is conditional on the value \(z\) takes.
• Finally, \(dz\) moves within the straightforward bounds of \(1\) to \(4\), covering the range of interest for \(z\).

Shifting limits can reframe the approach to problem-solving, letting you integrate more efficiently. Always confirm how the variables interconnect; it's crucial for correctly setting these boundaries.

Calculus Problem Solving

Problem-solving in calculus demands a structured approach. Often you need to:

• Visualize the problem, even sketch the regions if applicable, to grasp how it all fits together spatially.
• Reorganize the problem through changes like the order of integration, a key here as seen in transforming the sequence to \(dy\), \(dx\), \(dz\).
• Use substitution effectively, such as shifting from \(y\) to another variable to simplify the integration as observed with \(u = \sqrt{yz}\) in our case.
• Manage each integral independently before combining them back through sequential integration to extract the final result.

Mastering calculus problem solving is like uncovering layers, one step at a time. Each layer must be correctly peeled and examined, building towards the solution. With practice, this structured order becomes intuitive.