/*! 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 29 Evaluate the following integrals... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate the following integrals using polar coordinates. Assume \((r, \theta)\) are polar coordinates. A sketch is helpful. $$\iint_{R} \sqrt{x^{2}+y^{2}} d A ; R=\left\\{(x, y): 1 \leq x^{2}+y^{2} \leq 4\right\\}$$

Short Answer

Expert verified
Short Answer: The double integral of the function \(\sqrt{x^{2}+y^{2}}\) over the region R given by \(1 \leq x^{2}+y^{2} \leq 4\) is equal to \(\frac{14\pi}{3}\).

Step by step solution

01

Understand the geometry of the region R

R consists of the points \((x, y)\) satisfying \(1 \leq x^{2}+y^{2} \leq 4\). In polar coordinates, \(x^{2}+y^{2}=r^{2}\). Therefore, the region R can be written in terms of polar coordinates as the set of points \((r, \theta)\) such that: $$1 \leq r^{2} \leq 4$$ Taking square root in both sides: $$1 \leq r \leq 2$$ Since there is no restriction on the angle θ, it means that θ can range between 0 and 2π. Therefore, the limits for r and θ are: $$ 1\leq r\leq 2$$ $$0\leq\theta\leq2\pi$$
02

Convert the integrand to polar coordinates

Recall that the area element dA in Cartesian coordinates is dx dy, and in polar coordinates, it is rdr dθ. Also, the integrand function is \(\sqrt{x^{2}+y^{2}}\), which can be written in terms of r as: $$\sqrt{x^{2}+y^{2}}=\sqrt{r^{2}}=r$$ So, the double integral can be written in polar coordinates as: $$\iint_{R} \sqrt{x^{2}+y^{2}} dA = \int_{0}^{2\pi} \int_{1}^{2} r\cdot rdr d\theta$$
03

Evaluate the integral with respect to r

First, we'll integrate the inside integral with respect to r: $$\int_{1}^{2} r\cdot rdr = \int_{1}^{2} r^{2} dr = \frac{1}{3}r^{3}\Big|_1^2 = \frac{8}{3}-\frac{1}{3}=\frac{7}{3}$$
04

Evaluate the integral with respect to θ

Now we'll plug this result into the outside integral and integrate with respect to θ: $$\int_{0}^{2\pi} \left(\frac{7}{3}\right) d\theta = \frac{7}{3} \int_{0}^{2\pi}d\theta = \frac{7}{3}\left(\theta\Big|_0^{2\pi}\right) = \frac{7}{3}\cdot2\pi$$
05

Final result

So, the final result for the double integral is: $$\iint_{R} \sqrt{x^{2}+y^{2}} dA = \frac{14\pi}{3}$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Polar Coordinates
Polar coordinates offer a two-dimensional coordinate system where each point on a plane is determined by a distance from a reference point and an angle from a reference direction. This system is very useful in scenarios where the relationship between two points is easier to express with angles and distances, such as in the evaluation of integrals over circular regions.

In polar coordinates, the reference point is called the pole (akin to the origin in Cartesian coordinates), and the reference direction is usually the positive x-axis. A point is then denoted as \(r, \theta\), where \(r\) is the radial distance from the pole, and \(\theta\) is the angle measured in radians from the reference direction.

For example, Cartesian coordinates \(x, y\) can be converted to polar coordinates where \(x = r\cos\theta\) and \(y = r\sin\theta\). This is particularly helpful when dealing with regions bounded by circles or other radial symmetries as it simplifies the computation of areas and lengths.
Double Integral
A double integral allows us to compute the volume under a surface over a two-dimensional region. It is an extension of the concept of a single integral to two dimensions. When you integrate a function of two variables \(f(x, y)\) over a region \(R\) in the xy-plane, you're essentially summing up the values of \(f\) at all the points in \(R\), multiplied by an infinitesimal area element \(dA\).

The double integral of \(f\) over \(R\) is denoted as \(\iint_{R} f(x, y) dA\). To compute a double integral, we often change the order of integration or change the coordinates to simplify the integration process. In cases with radial symmetry, we switch to polar coordinates, which transforms the double integral into an iterated integral with two separate integrations, one for \(r\) and one for \(\theta\).
Area Element in Polar Coordinates
While the area element in Cartesian coordinates is straightforward, given by \(dA = dx dy\), the area element in polar coordinates reflects the geometry of the polar system. Since a small change in the angle \(\theta\) and the radius \(r\) sweeps out a small sector of a circle, the infinitesimal element of area, often represented as \(dA\), is given by \(r dr d\theta\).

The presence of \(r\) in the area element reflects the fact that the same change in \(\theta\) results in a larger arc length for larger values of \(r\). This means that the area is not just the product of the differentials (as in Cartesian coordinates), but also involves the variable \(r\). When setting up integrals in polar coordinates, it's essential to include this \(r\) factor in the integrand to ensure correct calculation of areas and volumes.
Polar Conversion of Integrand
When converting from Cartesian coordinates to polar coordinates for the purpose of evaluating an integral, the integrand and the limits of integration must both be converted. The integrand conversion involves substituting \(x = r\cos\theta\) and \(y = r\sin\theta\) into the original Cartesian integrand. Additionally, factors specific to polar coordinates, such as the \(r\) that arises from the polar area element, must be included in the new integrand.

For instance, the integral of \(\sqrt{x^{2}+y^{2}} dA\) over a circular region in Cartesian coordinates translates to \(\int\int r\cdot r dr d\theta\) in polar coordinates, with \(r\) accounting both for conversion of the integrand and the area element. The limits of integration are determined by the bounds of \(r\) and \(\theta\) for the particular region being integrated over, ensuring that the entire region is scanned exactly once. This method not only simplifies calculations but can also unveil symmetries that are not obvious in Cartesian form.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Improper integrals Improper integrals arise in polar coordinates when the radial coordinate r becomes arbitrarily large. Under certain conditions, these integrals are treated in the usual way: $$\int_{\alpha}^{\beta} \int_{a}^{\infty} g(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} g(r, \theta) r d r d \theta$$ Use this technique to evaluate the following integrals. $$\int_{0}^{\pi / 2} \int_{1}^{\infty} \frac{\cos \theta}{r^{3}} r d r d \theta$$

Consider the following two- and three-dimensional regions with variable dimensions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A solid is enclosed by a hemisphere of radius \(a\). How far from the base is the center of mass?

Gravitational field due to spherical shell A point mass \(m\) is a distance \(d\) from the center of a thin spherical shell of mass. \(M\) and radius \(R\). The magnitude of the gravitational force on the point mass is given by the integral $$F(d)=\frac{G M m}{4 \pi} \int_{0}^{2 \pi} \int_{0}^{\infty} \frac{(d-R \cos \varphi) \sin \varphi}{\left(R^{2}+d^{2}-2 R d \cos \varphi\right)^{3 / 2}} d \varphi d \theta$$ where \(G\) is the gravitational constant. a. Use the change of variable \(x=\cos \varphi\) to evaluate the integral and show that if \(d>R,\) then \(F(d)=G M m / d^{2},\) which means the force is the same as it would be if the mass of the shell were concentrated at its center. b. Show that if \(d

If possible, write an iterated integral in cylindrical coondinates of a function \(g(r, \theta, z)\) for the following regions in the specified orders. Sketch the region of integration. The solid outside the cylinder \(r=1\) and inside the sphere \(\rho=5\) for \(z \geq 0,\) in the orders \(d z d r d \theta, d r d z d \theta,\) and \(d \theta d z d r\).

Areas of circles Use integration to show that the circles \(r=2 a \cos \theta\) and \(r=2 a \sin \theta\) have the same area, which is \(\pi a^{2}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.