Problem 20 - University Calculus: Early Transcendentals

Question

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 integral evaluates to \( \frac{\pi}{4} \).

Step 1: Understanding the bounds of integration

The integral \( \int_{0}^{\pi} \int_{0}^{\sin x} y \, dy \, dx \) indicates that for each fixed \( x \) in the range \( 0 \leq x \leq \pi \), \( y \) varies from \( 0 \) to \( \sin x \). This implies that we should visualize the region as bounded by \( y = \sin x \), \( y = 0 \), \( x = 0 \), and \( x = \pi \).

Step 2: Sketching the region of integration

The region of integration is a region in the xy-plane. First, draw the graph of \( y = \sin x \) from \( x = 0 \) to \( x = \pi \). This is a sinusoidal curve starting and ending at the x-axis with a peak at \( x = \frac{\pi}{2} \). The region is below this curve and above the x-axis between \( x = 0 \) and \( x = \pi \).

Step 3: Evaluating the inner integral over y

To evaluate the inner integral \( \int_{0}^{\sin x} y \, dy \), we integrate \( y \) with respect to \( y \). The antiderivative of \( y \) is \( \frac{y^2}{2} \). Evaluate this from \( 0 \) to \( \sin x \), yielding: \[ \int_{0}^{\sin x} y \, dy = \frac{(\sin x)^2}{2}. \]

Step 4: Evaluating the outer integral over x

Now, evaluate the outer integral: \( \int_{0}^{\pi} \frac{(\sin x)^2}{2} \, dx \). Simplify by factoring out \( \frac{1}{2} \), the integral becomes \( \frac{1}{2} \int_{0}^{\pi} \sin^2 x \, dx \). Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to facilitate integration: \[ \frac{1}{2} \int_{0}^{\pi} \frac{1 - \cos 2x}{2} \, dx = \frac{1}{4} \int_{0}^{\pi} (1 - \cos 2x) \, dx. \]

Step 5: Computing the definite integral with trigonometric identity

Split the integral \( \frac{1}{4} \int_{0}^{\pi} (1 - \cos 2x) \, dx \) into two parts and compute each part separately: \[ \frac{1}{4} \left( \int_{0}^{\pi} 1 \, dx - \int_{0}^{\pi} \cos 2x \, dx \right). \] The first part evaluates to \( \int_{0}^{\pi} 1 \, dx = \pi \). The second part evaluates using the antiderivative of \( -\sin 2x \), yielding \( \frac{1}{2} \sin 2x \) which evaluates to zero over \( 0 \) to \( \pi \). Thus, the integral becomes \( \frac{1}{4} (\pi - 0) = \frac{\pi}{4}. \)

Key Concepts

Region of Integration

The region of integration in a double integral helps to define the area in the plane where the integration is performed. In this problem, the bounds are given as

• For the variable \( x \), the interval is from \( 0 \) to \( \pi \).
• For each fixed value of \( x \), the variable \( y \) ranges from \( 0 \) to \( \sin x \).

This indicates that the region of interest lies under the sinusoidal curve \( y = \sin x \), above the x-axis, and between \( x = 0 \) and \( x = \pi \).
When visualizing this, it can be helpful to sketch the curve \( y = \sin x \) across the specified range. You will see how the wave starts at the origin \((0,0)\), peaks at \( x = \frac{\pi}{2} \), and returns to zero at \( x = \pi \). The area of integration is a two-dimensional space under this curve.
The primary goal in sketching the region is to provide a visual context that underlines where \( y \) and \( x \) interact, aiding in understanding and solving the integral.

Sinusoidal Curve

Sinusoidal curves emerge from the sine function, commonly known for their wave-like shape. The function \( y = \sin x \) oscillates between its maximum value of 1 and minimum value of -1.
In this exercise, we focus on one positive arch of the sine function, which spans from \( x=0 \) to \( x=\pi \).

• At \( x = 0 \), \( y = \sin 0 = 0 \).
• At \( x = \frac{\pi}{2} \), \( y = \sin \frac{\pi}{2} = 1 \).
• At \( x = \pi \), \( y = \sin \pi = 0 \).

The particular characteristic we seek in such curves for this integral problem is the bounding nature from above the integration region. This curve dictates the variable limit for \( y \) as it stretches over the x-axis, forming what visually resembles a hill.
Understanding this aspect of sinusoidal functions can significantly ease handling integration tasks that include trigonometric-bound limits. They provide an intuitive north star on how the function behaves and intersects with axes.

Trigonometric Identities

Trigonometric identities are fundamental tools in calculus, providing pathways to simplify and solve integrals that involve trigonometric functions. Several key identities can transform complex integrals into simpler, computable forms.
In the case of double integrals with trigonometric-dependent regions, one helpful identity is:

• \( \sin^2 x = \frac{1 - \cos 2x}{2} \)

This particular identity allows us to convert an integral of \( \sin^2 x \) — which might appear challenging directly — into a more manageable expression.
For example, integrating \( \sin^2 x \) over \( x \) from 0 to \( \pi \), we can apply the substitution to

• \( \int_{0}^{\pi} \sin^2 x \, dx = \int_{0}^{\pi} \frac{1 - \cos 2x}{2} \, dx \).
• Further simplifying this is done by breaking it into two separate integrals: \( \int_{0}^{\pi} 1 \, dx \) and \( \int_{0}^{\pi} \cos 2x \, dx \).

Knowing such trigonometric identities does not only make integrals appear less intimidating, but also accelerates the computation process by reducing complexity.