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
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2c370c8-f220-47ec-b561-9f6a07b2c79a%2F9c644018-6583-408f-933a-69d8bb8dfd9f%2Fanrn8up_processed.png&w=3840&q=75)
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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