Problem 36 - Thomas Calculus

Question

Evaluate \(\int_{-1}^{1} \int_{0}^{\pi / 2} x \sin \sqrt{y} d y d x\)

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Interpret the Integral

The given integral is a double integral of the function \( x \sin \sqrt{y} \) over a rectangular region in the \(xy\)-plane. The bounds for \(x\) are from \(-1\) to \(1\) and for \(y\) are from \(0\) to \(\pi/2\). This means we first integrate with respect to \(y\) and then with respect to \(x\).

Step 2: Integrate with Respect to y

The inner integral is \( \int_{0}^{\pi / 2} x \sin \sqrt{y} \, dy \). Treating \(x\) as a constant, the integral becomes \( x \int_{0}^{\pi / 2} \sin \sqrt{y} \, dy \). Let \( u = \sqrt{y} \), then \( du = \frac{1}{2\sqrt{y}} dy \), implying \( dy = 2u \, du \). The limits of \(y\) transform from \(0\) to \(\pi/2\) as \(u\) transforms from \(0\) to \(\sqrt{\pi/2}\). Substitute into the integral: \( 2x \int_{0}^{\sqrt{\pi/2}} u \sin u \, du \).

Step 3: Evaluate the Substituted Integral

Use integration by parts for the integral \( \int u \sin u \, du \). Let \( v = u \) and \( dw = \sin u du \) so that \( dv = du \) and \( w = -\cos u \). Integration by parts gives:\( uv|_{0}^{\sqrt{\pi/2}} - \int_{0}^{\sqrt{\pi/2}} v \, dw \). Compute it as:\(-u \cos u |_{0}^{\sqrt{\pi/2}} + \int_{0}^{\sqrt{\pi/2}} \cos u \, du \).Thus, it becomes:\( - \left( \sqrt{\pi/2} \cos \sqrt{\pi/2} - 0 \right) + \left( \sin u |_{0}^{\sqrt{\pi/2}} \right)\).Finally, this results in:\( -\sqrt{\pi/2} \cos \sqrt{\pi/2} + \sin \sqrt{\pi/2} \).

Step 4: Integrate Result with Respect to x

Now plug this result into the integral with respect to \(x\):\( 2 \int_{-1}^{1} x \left(-\sqrt{\pi/2} \cos \sqrt{\pi/2} + \sin \sqrt{\pi/2} \right) dx \).Separate and evaluate each term separately:\( 2 \left(-\sqrt{\pi/2} \cos \sqrt{\pi/2} \int_{-1}^{1} x \, dx + \sin \sqrt{\pi/2} \int_{-1}^{1} x \, dx \right) \).Both integrals of \(x\) over \([-1, 1]\) result in zero. Hence each term evaluates to zero.

Step 5: Conclusion

Since both terms evaluated to zero, the value of the double integral \(\int_{-1}^{1} \int_{0}^{\pi / 2} x \sin \sqrt{y} \, dy \, dx\) is zero. This is because the function \(x\) is odd, integrating over symmetric bounds which results in cancellation.

Key Concepts

Integration by Parts

Integration by parts is a crucial technique for solving integrals where the integrand is a product of two functions. It is based on the formula:

• \( \int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx \)

In the provided exercise, integration by parts is used to handle the integral \( \int u \sin u \, du \). We choose \( u = u \) and \( dv = \sin u \, du \), leading to \( du = du \) and \( v = -\cos u \). By substituting into the integration by parts formula, we can break down the problem
into simpler integrals that are easier to evaluate. Finding appropriate parts for \( u \) and \( dv \) is key, often simplifying the integration process for more complex problems.
Using integration by parts requires practice to choose the functions wisely so as to simplify rather than complicate the integral.

Coordinate Transformation

A coordinate transformation can simplify the process of integration, particularly when dealing with complex integrands or limits. In the exercise,
a substitution was made with \( u = \sqrt{y} \). By transforming \( y \) into \( u \), the integral became easier to handle.

• \( u = \sqrt{y} \)
• \( du = \frac{1}{2\sqrt{y}} \, dy \)
• \( dy = 2u \, du \)

The limits were also changed from \( y = 0 \) to \( y = \pi/2 \) into \( u = 0 \) to \( u = \sqrt{\pi/2} \). This coordinate transformation allowed the integral to be expressed in terms of \( u \),
which was more manageable and straightforward to evaluate.Choosing the right substitution can simplify limits and the integrand, thus making the integration process smoother.

Odd Function Symmetry

Integrating odd functions over symmetric intervals can significantly simplify computations, often leading the integral to evaluate to zero. By definition:

• A function \( f(x) \) is odd if \( f(-x) = -f(x) \).

In such cases, symmetrical integration bounds like \([-a, a]\) result in cancellation. This is because the positive area on one side of the origin
matches the negative area on the other, leading to a net zero result. For this exercise, the function \( x \sin \sqrt{y} \) is integrated over \( x \) from -1 to 1, recognizing that:

• \( \int_{-1}^{1} x \, dx = 0 \)

Utilizing odd function symmetry is helpful in double integrals, saving time and computation when integrating across symmetric intervals.

Rectangular Region Integration

Double integrals over a rectangular region involve integration along two dimensions, typically with respect to \( y \) first and then \( x \), or vice versa. The region can be described by the bounds of integration for each variable. In this exercise:

• \( x \) bounds: \([-1, 1]\)
• \( y \) bounds: \([0, \pi/2]\)

The process involved first integrating with respect to \( y \), treating \( x \) as a constant, and then integrating the resulting expression with respect to \( x \).
By setting definite limits, the integration confines to a specific region in the Cartesian plane.This technique divides a complicated double integral into manageable parts,
allowing each to be solved systematically.