Problem 29 - University Calculus: Early Transcendentals

Question

Find the volume of the region bounded above by the surface \(z=2 \sin x \cos y\) and below by the rectangle \(R: 0 \leq x \leq \pi / 2\) \(0 \leq y \leq \pi / 4\).

Step-by-Step Solution

Answer

The volume is \(\sqrt{2}\).

Step 1: Understand the Region and the Surface

We are given a surface defined by the equation \(z = 2 \sin x \cos y\) and the bounds given by the rectangle \(R: 0 \leq x \leq \frac{\pi}{2}, 0 \leq y \leq \frac{\pi}{4}\). This means we need to calculate the volume under the surface within these limits.

Step 2: Set Up the Double Integral

To find the volume, we set up a double integral to evaluate the volume under the surface over the region \(R\). The volume \(V\) can be calculated as \(V = \int_0^{\pi/2} \int_0^{\pi/4} 2 \sin x \cos y \, dy \, dx\).

Step 3: Evaluate the Inner Integral

First, we calculate the inner integral with respect to \(y\):\[\int_0^{\pi/4} 2 \sin x \cos y \, dy = 2 \sin x \left[ \sin y \right]_0^{\pi/4}.\]Evaluating, we have:\[2 \sin x \left( \sin \frac{\pi}{4} - \sin 0 \right) = 2 \sin x \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \sin x.\]

Step 4: Evaluate the Outer Integral

Next, we calculate the outer integral with respect to \(x\):\[\int_0^{\pi/2} \sqrt{2} \sin x \, dx = \sqrt{2} \left[ -\cos x \right]_0^{\pi/2}.\]Evaluating gives:\[\sqrt{2} \left( -\cos \frac{\pi}{2} + \cos 0 \right) = \sqrt{2} \left( 0 + 1 \right) = \sqrt{2}.\]

Step 5: Conclusion

The volume of the region bounded under the surface and over the rectangle is \(\sqrt{2}\).

Key Concepts

Volume Calculation

When we want to calculate the volume of a region bounded by a surface, we turn to double integrals. Double integrals help us find volumes under surfaces over specified regions. In this exercise, the surface is described by the function \(z = 2 \sin x \cos y\), which varies with both \(x\) and \(y\).
By integrating this function over the defined region, we can find the volume beneath the surface but above the rectangle in the \(xy\)-plane. The double integral

• breaks down the problem of finding volume into simpler steps,
• first evaluating the inner integral for one variable,
• then evaluating the outer integral for the remaining variable.

In our case, the bounds \(0 \leq x \leq \frac{\pi}{2}\) and \(0 \leq y \leq \frac{\pi}{4}\) define a rectangular region over which we integrate. This method allows the calculation of the volume even for non-standard shapes, like our given surface.

Surface Integration

Surface integration involves evaluating functions over a bounded region to calculate properties like volume, area, or other physical properties. In our solution, the function \(z = 2 \sin x \cos y\) represents the surface we integrate over.
The process starts with setting up the double integral, which gives a framework to systematically solve the problem. The volume in this scenario is found by:

• integrating first with respect to \(y\) over its bounds,
• which simplifies then to an expression in terms of \(x\).

The simplified integral of \(\int_0^{\pi/4} 2 \sin x \cos y \, dy\) becomes \(\sqrt{2} \sin x\). We continue with the next step by resolving the remaining integral in terms of \(x\), which is now a simpler task.
Eventually, the entire integration process through both coordinates helps us find \(\sqrt{2}\) as the volume of the entire region.

Region Boundaries

Defining the region boundaries is crucial for setting up a correct double integral. In our exercise, the boundaries define a rectangle in the \(xy\)-plane. The limits for \(x\) are from \(0\) to \(\frac{\pi}{2}\) and for \(y\) from \(0\) to \(\frac{\pi}{4}\). These set the limits of our integrals.
Understanding these limits:

• can involve visualizing the area in the coordinate plane,
• ensures we only consider the values contributing to our volume calculation.

The bounded region essentially acts as the base of our volume.
Think of the process as stacking infinitesimally thin slices of area, stretching up to meet the surface \(z = 2 \sin x \cos y\). Each slice contributes a small volume that sums up via integration to give the entire volume under the surface but over this rectangular region.