Question

Use a triple integral to compute the volume of the following regions. The solid common to the cylinders \(z=\sin x\) and \(z=\sin y\) over the square \(R=\\{(x, y): 0 \leq x \leq \pi, 0 \leq y \leq \pi\\}\) (The figure shows the cylinders, but not the common region.)

Short answer

Answer: The volume of the region is \(3\pi^2\).

Step-by-step solution

Set up the triple integral

First, we need to set up the triple integral that represents the volume of the region we want to find. To do this, we will take the volume element \(dz\, dy\, dx\). The function to integrate is simply 1, but we need to find the appropriate limits of integration for each of the variables.

Finding the limits of integration for x and y

As given, x and y are already bounded by the square R. Therefore, the limits of integration for x and y are: 0 \(\leq x \leq \pi\) 0 \(\leq y \leq \pi\)

Finding the limits of integration for z

Now we need to find the limits of integration for z. Since z is the height of the solid, we can analyze the region formed by the intersection of the two cylinders. From the given functions \(z=\sin x\) and \(z=\sin y\), we can deduce that the limits of integration for z are: \(\max\{0, \sin x - \sin y\} \leq z \leq \min\{\sin x, \sin y\}\)

Setting up the triple integral

Now that we have the limits of integration, we can set up the triple integral: \(\int_{0}^{\pi}\int_{0}^{\pi}\int_{\max\{0, \sin x - \sin y\}}^{\min\{\sin x, \sin y\}} 1\, dz\, dy\, dx\)

Evaluate the inner integral

First, we will evaluate the z integral: \(\int_{0}^{\pi}\int_{0}^{\pi}(\min\{\sin x, \sin y\} - \max\{0, \sin x - \sin y\})\, dy\, dx\)

Split the integral into cases

Now, we will split the integral into cases based on the relationships between \(\sin x\) and \(\sin y\): Case 1: \(\sin x \leq \sin y\) \(\int_{0}^{\pi}\int_{0}^{\pi}(\sin x - 0)\, dy\, dx\) Case 2: \(\sin x \geq \sin y\) \(\int_{0}^{\pi}\int_{0}^{\pi}(\sin y - (\sin x - \sin y))\, dy\, dx\)

Evaluate the remaining integrals

Now we will evaluate the remaining integrals for each case: Case 1: \(\int_{0}^{\pi}(\pi\sin x)\, dx = \boxed{\pi^2}\) Case 2: \(\int_{0}^{\pi} 2 (\pi\sin y)\, dy\, dx = \boxed{2\pi^2}\)

Add the results from both cases

Finally, we will add the results of both cases to get the total volume: Total Volume = \(\pi^2 + 2\pi^2 = \boxed{3\pi^2}\)