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Chapter 9: Problem 53

Using symmetry, \(I_{x}=2 \int_{0}^{\pi / 2} \int_{0}^{\cos x} k y^{2} d y d x=\left.2 k \int_{0}^{\pi / 2} \frac{1}{3} y^{3}\right|_{0} ^{\cos x} d x=\frac{2}{3} k \int_{0}^{\pi / 2} \cos ^{3} x d x\) \(=\frac{2}{3} k \int_{0}^{\pi / 2} \cos x\left(1-\sin ^{2} x\right) d x=\left.\frac{2}{3} k\left(\sin x-\frac{1}{3} \sin ^{3} x\right)\right|_{0} ^{\pi / 2}=\frac{4}{9} k\).

Short Answer

Expert verified
The value of the integral is \(\frac{4}{9} k\).

Step by step solution

01

Understanding the Symmetry

We're given an integral that utilizes symmetry to simplify the initial calculation. The integral represents an area or volume within certain bounds, simplified by evaluating only half of the domain and then multiplying the result by 2 to account for symmetry.
02

First Integration with respect to y

The expression begins with \(2 \, \int_{0}^{\pi/2} \int_{0}^{\cos x} k y^{2} dy \, dx\). We perform the integration first with respect to \(y\). This results in \(\left. 2k \left( \frac{1}{3} y^{3} \right) \right|_{0}^{\cos x}\).
03

Substitute the Limits for y

Substitute \(y = \cos x\) and \(y = 0\) into the expression \(\frac{1}{3} y^3\). This results in \(\frac{1}{3} (\cos x)^3\), which simplifies the integral to \(\frac{2}{3} k \int_{0}^{\pi/2} \cos^{3} x \, dx\).
04

Simplify the Integral with Trigonometric Identity

The expression \(\cos^{3} x\) is rewritten using the identity \(\cos x \cdot (1 - \sin^{2} x)\). This gives \(\frac{2}{3} k \int_{0}^{\pi/2} \cos x (1 - \sin^{2} x) \, dx\).
05

Substitute for cos x and Integrate

The expression simplifies to \(\frac{2}{3} k \left( \int_{0}^{\pi/2} \cos x \, dx - \int_{0}^{\pi/2} \cos x \sin^{2} x \, dx \right)\). Each integral is evaluated separately.
06

Evaluate Boundary Conditions

After calculating the integrals, we substitute the boundary conditions, \(x = 0\) and \(x = \pi/2\), into \(\sin x - \frac{1}{3} \sin^{3} x\), resulting in \(\frac{2}{3} k \left( 1 - 0 \right)\).
07

Final Calculation

The final expression from the boundary evaluations gives us \(\frac{4}{9} k\), which is the solution of the given problem.

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Key Concepts

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

Symmetry in Integration
When dealing with integrals, symmetry can often be utilized as a powerful tool to simplify complex problems. In the given problem, symmetry comes into play when calculating the integral over a symmetric region. Consider an area that is symmetrical about an axis; instead of calculating the integral over the entire area, we can calculate it over half the domain, then multiply the result by 2. This approach reduces the computation effort while maintaining accuracy. For instance, the integral provided is evaluated only over the interval from 0 to \( \pi/2 \), and, thanks to symmetry, multiplying by 2 accounts for the full interval from 0 to \( \pi \). This clever use of symmetry is particularly useful in physics and engineering, where symmetrical structures are common.
Trigonometric Identities
Trigonometric identities are mathematical tools that express relationships between trigonometric functions. They are invaluable in calculus, specifically when simplifying integrals involving trigonometric expressions. In this exercise, the identity \( \cos^3 x = \cos x (1 - \sin^2 x) \) is used. By rewriting \( \cos^3 x \) in terms of simpler components, we can break down the complex integral into more manageable parts. This strategy simplifies manipulations and integrations, helping to reach the solution more efficiently. Familiarity with various trigonometric identities strengthens problem-solving skills and opens up different avenues for tackling problems in calculus.
Integral Boundaries
Integral boundaries define the limits within which an integral is evaluated. They are critical as they determine the area or volume being considered. In the integral \( \int_{0}^{\pi/2} \), the boundaries \(0\) and \(\pi/2\) specify the region of interest on the x-axis. These limits result from examining the geometry of the problem, often linked to curves or physical constraints. Choosing the correct boundaries is vital as it ensures the integral accurately represents the problem at hand. Moreover, when evaluating definite integrals, substituting these boundary values reveals the net area or volume, providing meaningful interpretations of the result.
Area Under Curves
The area under a curve is a fundamental concept in integral calculus. It corresponds to the quantitative measure of the space beneath a curve on a graph, between two limits. In practice, evaluating an integral like \( \int_{0}^{\pi/2} \cos^3 x \, dx \) represents calculating the area under the curve of \( \cos^3 x \) from 0 to \( \pi/2 \). This is important in many fields, such as physics, for computing work done by variable forces, and in economics, for finding consumer surplus. Understanding how to compute the area under curves allows for practical applications of calculus in real-world situations, and aids in visualizing the accumulative measures represented by integrals.

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Most popular questions from this chapter

$$\begin{aligned}\mathbf{B}+\mathbf{W} &=-\iint_{S} p \mathbf{n} d S+m \mathbf{g}=m \mathbf{g}-\iiint_{D} \nabla p d V=m \mathbf{g}-\iiint_{D} \rho \mathbf{g} d V=m \mathbf{g}-\left(\iiint_{D} \rho d V\right) \mathbf{g} \\ &=m \mathbf{g}-m \mathbf{g}=\mathbf{0}\end{aligned}$$

The density is \(\rho=k y .\) Using symmetry, \(I_{y}=2 \int_{0}^{1} \int_{0}^{1-x^{2}} k x^{2} y d y d x=\left.2 \int_{0}^{1} \frac{1}{2} k x^{2} y^{2}\right|_{0} ^{1-x^{2}} d x=k \int_{0}^{1} x^{2}\left(1-x^{2}\right)^{2} d x\) \(=k \int_{0}^{1}\left(x^{2}-2 x^{4}+x^{6}\right) d x=\left.k\left(\frac{1}{3} x^{3}-\frac{2}{5} x^{5}+\frac{1}{7} x^{7}\right)\right|_{0} ^{1}=\frac{8 k}{105}\).

\(\operatorname{div} \mathbf{F}=2(z-1) .\) Using cylindrical coordinates, $$\begin{aligned}\iint_{S} \mathbf{F} \cdot \mathbf{n} d S &=\iiint_{D} 2(z-1) V=\int_{0}^{2 \pi} \int_{0}^{4} \int_{1}^{5} 2(z-1) d z r d r d \theta=\left.\int_{0}^{2 \pi} \int_{0}^{4}(z-1)^{2}\right|_{1} ^{5} r d r d \theta \\\&=\int_{0}^{2 \pi} \int_{0}^{4} 16 r d r d \theta=\left.\int_{0}^{2 \pi} 8 r^{2}\right|_{0} ^{4} d \theta=128 \int_{0}^{2 \pi} d \theta=256 \pi\end{aligned}$$

\(I_{y}=\int_{0}^{1} \int_{x^{2}}^{\sqrt{x}} x^{4} d y d x=\left.\int_{0}^{1} x^{4} y\right|_{x^{2}} ^{\sqrt{x}} d x=\int_{0}^{1}\left(x^{9 / 2}-x^{6}\right) d x=\left.\left(\frac{2}{11} x^{11 / 2}-\frac{1}{7} x^{7}\right)\right|_{0} ^{1}=\frac{3}{77}\)

The surface is \(g(x, y, z)=x+y+z-6 . \quad \nabla g=\mathbf{i}+\mathbf{j}+\mathbf{k},|\nabla g|=\sqrt{3} ; \mathbf{n}=(\mathbf{i}+\mathbf{j}+\mathbf{k}) / \sqrt{3}\) \(\mathbf{F} \cdot \mathbf{n}=\left(e^{y}+e^{x}+18 y\right) / \sqrt{3} ; z_{x}=-1, z_{y}=-1, d S=\sqrt{1+1+1} d A=\sqrt{3} d A\) $$\begin{aligned} \mathrm{Flux} &=\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=\iint_{r}\left(e^{y}+e^{x}+18 y\right) d A=\int_{0}^{6} \int_{0}^{6-x}\left(e^{y}+e^{x}+18 y\right) d y d x \\ &=\left.\int_{0}^{6}\left(e^{y}+y e^{x}+9 y^{2}\right)\right|_{0} ^{6-x} d x=\int_{0}^{6}\left[e^{6-x}+(6-x) e^{x}+9(6-x)^{2}-1\right] d x \\ &=\left.\left[-e^{6-x}+6 e^{x}-x e^{x}+e^{x}-3(6-x)^{3}-x\right]\right|_{0} ^{6} \\ &=\left(-1+6 e^{6}-6 e^{6}+e^{6}-6\right)-\left(-e^{6}+6+1-648\right)=2 e^{6}+634 \approx 1440.86 \end{aligned}$$
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