Problem 6 - Thomas Calculus

Question

In Exercises \(1-10,\) sketch the region of integration and evaluate the integral. $$ \int_{0}^{\pi} \int_{0}^{\sin x} y d y d x $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{\pi}{4} \).

Step 1: Understand the Region of Integration

The integral \( \int_{0}^{\pi} \int_{0}^{\sin x} y \, dy \, dx \) represents an area in the xy-plane. The limits \( y = 0 \) to \( y = \sin x \) and \( x = 0 \) to \( x = \pi \) define a region. The curve \( y = \sin x \) bounds the region above, which means it lies below the sine function between \( x = 0 \) and \( x = \pi \).

Step 2: Sketch the Region of Integration

To plot the region of integration, draw the sine curve from \( x = 0 \) to \( x = \pi \). Then, mark the region from \( y = 0 \) to \( y = \sin x \). This forms a band under one full period of the sine wave from \( x = 0 \) to \( x = \pi \).

Step 3: Change the Order of Integration if Necessary

For this integral, it's straightforward to proceed with the order given. However, knowing how to set \( x \) and \( y \) if needed could be valuable: consider alternatives such as setting bounds differently if the integral seemed complex to evaluate directly.

Step 4: Evaluate the Inner Integral

Evaluate the inner integral with respect to \( y \):\[\int_{0}^{\sin x} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\sin x} = \frac{(\sin x)^2}{2} - 0 = \frac{\sin^2 x}{2}.\]

Step 5: Evaluate the Outer Integral

Substitute the result of the inner integral into the outer integral, and then evaluate:\[\int_{0}^{\pi} \frac{\sin^2 x}{2} \, dx.\]Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to simplify further.\[= \frac{1}{2} \int_{0}^{\pi} \frac{1 - \cos 2x}{2} \, dx = \frac{1}{4} \int_{0}^{\pi} (1 - \cos 2x) \, dx.\]Now separate into two integrals:\[= \frac{1}{4} \left[ \int_{0}^{\pi} 1 \, dx - \int_{0}^{\pi} \cos 2x \, dx \right].\]

Step 6: Calculate Each Integral

Evaluate each part:1. \( \int_{0}^{\pi} 1 \, dx = [x]_{0}^{\pi} = \pi - 0 = \pi. \)2. \( \int_{0}^{\pi} \cos 2x \, dx = \left[ \frac{\sin 2x}{2} \right]_{0}^{\pi} = \frac{\sin 2\pi}{2} - \frac{\sin 0}{2} = 0.\)

Step 7: Final Calculation

Substitute back:\[= \frac{1}{4} \times (\pi - 0) = \frac{\pi}{4}.\]Thus, the value of the double integral is \( \frac{\pi}{4} \).

Key Concepts

Region of Integration

When approaching a double integral, the first step is understanding the region of integration within the xy-plane. This involves identifying the area over which the function will be integrated. For the given problem, we are working with the integral \( \int_{0}^{\pi} \int_{0}^{\sin x} y \, dy \, dx \). This tells us that we are examining the space where \( x \) ranges from \( 0 \) to \( \pi \), while \( y \) varies from \( 0 \) to \( \sin x \) for each value of \( x \).

This forms a region under the sine curve, specifically between the x-axis and the sine wave from \( x = 0 \) to \( x = \pi \). To visualize this, plot \( y = \sin x \) and shade the area under the curve within the given x-bounds. Understanding this region is fundamental as it allows us to grasp what area of the plane is being considered in the integral calculation.

Order of Integration

The order of integration refers to the sequence in which integrations are performed in a double integral. For \( \int_{0}^{\pi} \int_{0}^{\sin x} y \, dy \, dx \), we start by integrating with respect to \( y \) first, followed by \( x \). This is because the limits of \( y \) are dependent on the value of \( x \).

Sometimes, it is beneficial to change the order, especially if simplifying the integral can make calculations easier. If the order is to be reversed, the integration boundaries need to be recalculated to reflect this change. For this particular example, sticking to the given order is simpler as it aligns directly with the region's boundaries. But if challenges in evaluating were present, rewriting the integral with \( x \) first could be a valid strategic move. The decision could hinge on which variable allows easier antiderivative computations based on function behavior and interval constraints.

Integral Evaluation

The integral evaluation process involves calculating the actual value of the double integral once the integration order and region are both established. Begin by evaluating the inner integral: we calculate \( \int_{0}^{\sin x} y \, dy \), with \( y \) as the variable. This results in \( \frac{(\sin x)^2}{2} \).

The next phase involves substituting this result back into the outer integral, \( \int_{0}^{\pi} \frac{\sin^2 x}{2} \, dx \).

To simplify, employ the trigonometric identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \). By using this identity, break down the integral into separate, more manageable parts: \( \int_{0}^{\pi} 1 \, dx \) and \( \int_{0}^{\pi} \cos 2x \, dx \). Evaluating these separately, the integral of \( 1 \, dx \) over the interval yields \( \pi \), while the integral of \( \cos 2x \, dx \) results in zero due to the periodic nature of the sine function. Ultimately, compiling these results gives the solution for the original double integral, \( \frac{\pi}{4} \). This step-by-step breakdown facilitates understanding of not just how values change, but why such evaluations matter in determining the integral's total area.