/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Sketch the region \(R=\left\\{(r... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the region \(R=\left\\{(r, \theta) \mid 1 \leq r \leq 2,-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\right\\}\), and evaluate \(\iint_{R} x d A\)

Short Answer

Expert verified
The integral evaluates to \(\frac{14}{3}.\)

Step by step solution

01

Understand the Region R

The region \[R=\left\{(r, \theta) \mid 1 \leq r \leq 2,-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}\right\}\]is a part of the polar coordinate system. It is the region between the circles with radii 1 and 2, where the angle \(\theta\) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). This corresponds to a wedge shape that opens in the first and fourth quadrants.
02

Convert the Integral

The integral \[\iint_{R} x \, dA\]in polar coordinates can be expressed as \[\iint_{R} (r \cos \theta) \, r \, dr \, d\theta\]because in polar coordinates, \(x = r \cos \theta\) and \(dA = r \, dr \, d\theta\).
03

Set Up the Double Integral

The double integral becomes \[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{1}^{2} r^2 \cos \theta \, dr \, d\theta.\]Here, \(r\) ranges from 1 to 2, and \(\theta\) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
04

Evaluate the Inner Integral

Evaluate \[\int_{1}^{2} r^2 \, dr\]to get \[\left[ \frac{r^3}{3} \right]_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.\]
05

Evaluate the Outer Integral

Now integrate the result:\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{7}{3} \cos \theta \, d\theta.\]Using the antiderivative of \(\cos \theta\), which is \(\sin \theta\), we have:\[\frac{7}{3} \left[ \sin \theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{7}{3} (1 - (-1)) = \frac{7}{3} \times 2 = \frac{14}{3}.\]
06

Finalize the Solution

The result of the integral, \[\iint_{R} x \, dA\] is \[\frac{14}{3}.\]This represents the value of the integral over the region \(R\) in the \((x, y)\)-plane.

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Key Concepts

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

Double Integral
A double integral is a fundamental concept in calculus, allowing us to compute the volume under a surface in a two-dimensional region. Think of it as extending the idea of summing up areas in one-dimensional calculus to two dimensions. Double integrals are essential for calculating aspects such as mass, volume, and area in higher dimensions.
To perform a double integral, you first need to define a region in your plane where the integration will occur. In our problem, this region is defined in polar coordinates, spanning certain intervals of radial and angular values.
The double integral is often denoted as \( \iint_R f(x, y) \, dA \), where \( R \) signifies the region of integration, and \( dA \) represents a small area element. Here, the goal was to evaluate \( \iint_{R} x \, dA \), with conversion to polar coordinates.
Polar Integration
Polar integration refers to the process of evaluating integrals where the region is described in terms of polar coordinates \((r, \theta)\) instead of Cartesian coordinates \((x, y)\). This conversion is particularly useful for regions like circles and their sectors.
To convert a Cartesian coordinate function into polar coordinates for integration, you need to use the relationships:
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
  • \( dA = r \, dr \, d\theta \)
This setup simplifies integration over circular regions or sectors.
In our task, this conversion transformed the integral \( \iint_{R} x \, dA \) into \( \iint_{R} (r \cos \theta) \, r \, dr \, d\theta \), leading to the simplified form \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{1}^{2} r^2 \cos \theta \, dr \, d\theta \). This structured approach allowed us to evaluate the integral efficiently.
Cosine Function
The cosine function \( \cos \theta \) represents how much the projection of a point on a circle in two dimensions lies on the x-axis. It is one of the basic trigonometric functions, characterized by its oscillating wave-like graph.
The cosine of an angle \( \theta \) is the x-coordinate of the corresponding point on the unit circle. Cosine values oscillate between -1 and 1, with specific key points such as:
  • \( \cos(0) = 1 \)
  • \( \cos(\frac{\pi}{2}) = 0 \)
  • \( \cos(\pi) = -1 \)
These properties are crucial when evaluating integrals that involve trigonometric terms.
In polar integration, the cosine function \( \cos \theta \) was applied to simplify and solve the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{7}{3} \cos \theta \, d\theta \). Its antiderivative is \( \sin \theta \), leading to an efficient computation of \( \frac{14}{3} \) as the final result.

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Most popular questions from this chapter

If the charge density at an arbitrary point \((x, y, z)\) of a solid \(E\) is given by the function \(\rho(x, y, z)\), then the total charge inside the solid is defined as the triple integral \(\iiint_{E} \rho(x, y, z) d V .\) Assume that the charge density of the solid \(E\) enclosed by the paraboloids \(x=5-y^{2}-z^{2}\) and \(x=y^{2}+z^{2}-5\) is equal to the distance from an arbitrary point of \(E\) to the origin. Set up the integral that gives the total charge inside the solid \(E\).

True or False? Justify your answer with a proof or a counterexample.The Jacobian of the transformation for \(x=u^{2}-2 v, y=3 v-2 u v\) is given by \(-4 u^{2}+6 u+4 v\).

Evaluate the triple integrals over the bounded region \(E\) of the form \(E=\left\\{(x, y, z) \mid g_{1}(y) \leq x \leq g_{2}(y), c \leq y \leq d, e \leq z \leq f\right\\}\) $$ \iiint_{E} x^{2} d V, \text { where } E=\left\\{(x, y, z) \mid 1-y^{2} \leq x \leq y^{2}-1,-1 \leq y \leq 1,1 \leq z \leq 2\right\\} $$

Let \(Q\) be a solid of constant density \(k\), where \(k>0\), that is located in the first octant, inside the circular cone \(x^{2}+y^{2}=9(z-1)^{2}\), and above the plane \(z=0 .\) Show that the moment \(M_{x y}\) about the \(x y\) -plane is the same as the moment \(M_{y z}\) about the \(x z\) -plane.

For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure). A radial function \(f\) is a function whose value at each point depends only on the distance between that point and the origin of the system of coordinates; that is, \(f(x, y)=g(r)\), where \(r=\sqrt{x^{2}+y^{2}} .\) Show that if \(f\) is a continuous radial function, then \(\iint_{D} f(x, y) d A=\left(\theta_{2}-\theta_{1}\right)\left[G\left(R_{2}\right)-G\left(R_{1}\right)\right]\), where \(G^{\prime}(r)=r g(r)\) and \((x, y) \in D=\left\\{(r, \theta) \mid R_{1} \leq r \leq R_{2}, 0 \leq \theta \leq 2 \pi\right\\}\), with \(0 \leq R_{1}

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