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Change the Cartesian integral to an equivalent polar integral, and then evaluate. 10 √100-x² S - 10 A. B. D. S √100-x² 100 101 200 101 400 101 100 201 π π π T 1 (1+x² + y²) ² dy dx

Change the Cartesian integral to an equivalent polar integral, and then evaluate. 10 √100-x² S - 10 A. B. D. S √100-x² 100 101 200 101 400 101 100 201 π π π T 1 (1+x² + y²) ² dy dx

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem Statement:** Change the Cartesian integral to an equivalent polar integral, and then evaluate. \[ \int_{-10}^{10} \int_{-\sqrt{100-x^2}}^{\sqrt{100-x^2}} \frac{1}{\left(1+x^2+y^2\right)^2} \, dy \, dx \] **Multiple Choice Options:** A. \(\frac{100}{101}\pi\) B. \(\frac{200}{101}\pi\) C. \(\frac{400}{101}\pi\) D. \(\frac{100}{201}\pi\) **Correct Answer:** - Option A: \(\frac{100}{101}\pi\) **Explanation:** To solve the given integral, convert the Cartesian coordinates to polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \). The given integral's region is a circle of radius 10 centered at the origin, described in polar coordinates by \( 0 \leq r \leq 10 \) and \( 0 \leq \theta \leq 2\pi \). The function \( \frac{1}{\left(1+x^2+y^2\right)^2} \) transforms to \( \frac{1}{\left(1+r^2\right)^2} \) in polar coordinates. The differential area element \( dy \, dx \) converts to \( r \, dr \, d\theta \). Thus, the polar integral becomes: \[ \int_{0}^{2\pi} \int_{0}^{10} \frac{r}{(1+r^2)^2} \, dr \, d\theta \] Evaluate this integral with respect to \( r \) and \( \theta \) to obtain the result.
Transcribed Image Text:**Problem Statement:** Change the Cartesian integral to an equivalent polar integral, and then evaluate. \[ \int_{-10}^{10} \int_{-\sqrt{100-x^2}}^{\sqrt{100-x^2}} \frac{1}{\left(1+x^2+y^2\right)^2} \, dy \, dx \] **Multiple Choice Options:** A. \(\frac{100}{101}\pi\) B. \(\frac{200}{101}\pi\) C. \(\frac{400}{101}\pi\) D. \(\frac{100}{201}\pi\) **Correct Answer:** - Option A: \(\frac{100}{101}\pi\) **Explanation:** To solve the given integral, convert the Cartesian coordinates to polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \). The given integral's region is a circle of radius 10 centered at the origin, described in polar coordinates by \( 0 \leq r \leq 10 \) and \( 0 \leq \theta \leq 2\pi \). The function \( \frac{1}{\left(1+x^2+y^2\right)^2} \) transforms to \( \frac{1}{\left(1+r^2\right)^2} \) in polar coordinates. The differential area element \( dy \, dx \) converts to \( r \, dr \, d\theta \). Thus, the polar integral becomes: \[ \int_{0}^{2\pi} \int_{0}^{10} \frac{r}{(1+r^2)^2} \, dr \, d\theta \] Evaluate this integral with respect to \( r \) and \( \theta \) to obtain the result.
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