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Chapter 14: Problem 32

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

Short Answer

Expert verified
The integral evaluates to 0.

Step by step solution

01

Set Up the Double Integral

The given integral is a double integral, \[ \int_{-1}^{1} \int_{0}^{\pi / 2} x \sin \sqrt{y} \, dy \, dx .\] The outer integral is from \(-1\) to \(1\) with respect to \(x\), and the inner integral is from \(0\) to \(\pi/2\) with respect to \(y\). Here, \(x\) is treated as a constant when integrating with respect to \(y\).
02

Integrate with Respect to \(y\)

First, integrate the inner integral with respect to \(y\):\[ \int_{0}^{\pi / 2} x \sin \sqrt{y} \, dy. \]Since \(x\) is constant with respect to \(y\), we take it out of the integral:\[ x \int_{0}^{\pi / 2} \sin \sqrt{y} \, dy.\]Let \(u = \sqrt{y}\), so \(y = u^2\) and \(dy = 2u \, du\). When \(y = 0\), \(u = 0\), and when \(y = \pi/2\), \(u = \sqrt{\pi/2}\). The integral becomes:\[ x \int_{0}^{\sqrt{\pi/2}} \sin u \cdot 2u \, du. \]
03

Simplify the Integral with Substitution

Now, simplify the integral using the substitution:\[2x \int_{0}^{\sqrt{\pi/2}} u \sin u \, du.\]This is an integral that requires integration by parts. Let \(v = u\), \(dv = du\), and \(dw = \sin u \, du\), \(w = -\cos u\). Then:\[\int u \sin u \, du = -u \cos u + \int \cos u \, du.\]The integral of \(\cos u\) is \(\sin u\), so:\[\int u \sin u \, du = -u \cos u + \sin u\]evaluated from \(0\) to \(\sqrt{\pi/2}\).
04

Evaluate the Simplified Integral

Substitute back into the integral:\[2x \left[-u \cos u + \sin u\right]_{0}^{\sqrt{\pi/2}}.\]First, evaluate at \(u = \sqrt{\pi/2}\):\[= 2x \left(-\sqrt{\pi/2} \cos \sqrt{\pi/2} + \sin \sqrt{\pi/2}\right).\]Then evaluate at \(u = 0\):\[= 2x \left(0 \cdot \cos 0 + \sin 0\right) = 0.\]
05

Complete the Integral with Respect to \(x\)

Now integrate the resultant function with respect to \(x\):\[ \int_{-1}^{1} 2x \left(-\sqrt{\pi/2} \cos \sqrt{\pi/2} + \sin \sqrt{\pi/2}\right) \, dx.\]Since both \(-\sqrt{\pi/2} \cos \sqrt{\pi/2}\) and \(\sin \sqrt{\pi/2}\) are constants, the integral simplifies to:\[\left(-\sqrt{\pi/2} \cos \sqrt{\pi/2} + \sin \sqrt{\pi/2}\right) \int_{-1}^{1} 2x \, dx.\]Evaluate:\[ \int_{-1}^{1} 2x \, dx = 2 \left[\frac{x^2}{2}\right]_{-1}^{1} = 2 \left(1 - 1\right) = 0.\]
06

Conclusion

The result of the entire double integral is 0 because the integral of an odd function \(x\) multiplied by a constant, evaluated over symmetric limits \([-1, 1]\), results in 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by parts is a technique used to integrate products of functions. It is derived from the product rule for differentiation. The formula can be written as: \[ \int u\, dv = uv - \int v\, du \]To use this method, you must identify parts of the integral to label as \( u \) and \( dv \). Generally, choose \( u \) such that its derivative \( du \) is simpler, and select \( dv \) so that \( v \), its anti-derivative, is easy to find.
  • Apply the integration by parts in the given exercise when encountering the term \( u \sin{u} \).
  • Choosing \( u = u \) and \( dv = \sin{u} \, du \) simplifies the integration process.
  • This leads to the expression \(- u \cos{u} + \int \cos{u} \, du \).
Using integration by parts simplifies complex products into more manageable expressions, aiding in solving double integrals.
U-Substitution
U-substitution simplifies integrals by changing variables, enhancing ease of integration. It is similar to the substitution rule in differentiation. To apply u-substitution:
  • Select a function \( u \) such that \( du \) corresponds to a part of the integral.
  • Rewrite the integrand in terms of \( u \) and \( du \).
For our exercise, let \( u = \sqrt{y} \), then \( du = \frac{1}{2\sqrt{y}} \, dy \), which transforms \( dy \to 2u \, du \). This change of variables repositions the integration limits from \( y = 0 \) to \( y = \pi/2 \) into \( u = 0 \) to \( u = \sqrt{\pi/2} \).This method allows us to tackle potentially complicated integral expressions by simplifying the calculation through variable transformation.
Odd Functions
Odd functions are functions that satisfy the property \( f(-x) = -f(x) \).
  • Graphically, odd functions are symmetrical about the origin.
  • Examples include \( x \), \( \sin{x} \), and other types of functions.
In this exercise, the function \( 2x \) is odd since replacing \( x \) with \( -x \) changes the sign of the full term. When we integrate an odd function over symmetric intervals around zero \([-a, a]\), the result is zero because the positive and negative areas cancel each other out, resulting in zero net area.Utilizing the property of odd functions is helpful to quickly simplify computations within symmetric limits.
Symmetric Limits
Symmetric limits occur when the boundaries of an integral are equal in magnitude but opposite in sign, such as \([-1, 1]\). This symmetry plays a crucial role when dealing with certain types of functions.
  • For even functions, which satisfy \( f(-x) = f(x) \), integration over symmetric limits results in doubling the integral over half the interval \([0, a]\).
  • For odd functions, such as in the given exercise, integrating over symmetric limits results in a zero integral, as seen in \( \int_{-1}^{1} 2x \, dx \).
This concept allows for verification of results from an integral, enhancing understanding and enabling simplifications through recognizing patterns in integrable expressions and their limits.

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Most popular questions from this chapter

Evaluate the integral \(\iint_{R}\left(x^{2}+y^{2}\right)^{-2} d A,\) where \(R\) is the region inside the circle \(x^{2}+y^{2}=2\) for \(x \leq-1.\)

Minimizing a double integral What region \(R\) in the \(x y\) -plane minimizes the value of $$\iint_{R}\left(x^{2}+y^{2}-9\right) d A ?$$ Give reasons for your answer.

Use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. a. Plot the Cartesian region of integration in the \(x y\)-plane. b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for \(r\) and \(\theta\) c. Using the results in part (b), plot the polar region of integration in the \(r \theta\)-plane. d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility. $$\int_{0}^{1} \int_{0}^{x / 2} \frac{x}{x^{2}+y^{2}} d y d x$$

In polar coordinates, the average value of a function over a region \(R\) (Section 14.3 ) is given by $$\frac{1}{\operatorname{Area}(R)} \iint_{R} f(r, \theta) r d r d \theta$$ Find the average height of the hemispherical surface \(z=\sqrt{a^{2}-x^{2}-y^{2}}\) above the disk \(x^{2}+y^{2} \leq a^{2}\) in the \(x y\)-plane.

Let \(D\) be the region in the first octant that is bounded below by the cone \(\phi=\pi / 4\) and above by the sphere \(\rho=3 .\) Express the volume of \(D\) as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find \(V\).
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