Use polar coordinates to find the volume of the given solid. Below the cone z = V x2 + y2 and above the ring 1s x2 + y2 s 64
Use polar coordinates to find the volume of the given solid. Below the cone z = V x2 + y2 and above the ring 1s x2 + y2 s 64
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Use polar coordinates to find the volume of the given solid.
**Description of Solid:**
- The solid lies below the cone described by the equation:
\[
z = \sqrt{x^2 + y^2}
\]
- The solid is above the ring defined by the inequality:
\[
1 \leq x^2 + y^2 \leq 64
\]
**Instructions for Solution:**
To solve this problem, convert the Cartesian coordinates to polar coordinates. In polar coordinates, the relationships \(x = r\cos\theta\) and \(y = r\sin\theta\) are useful. The equation of the cone becomes \(z = r\), and the ring boundaries translate to \(1 \leq r^2 \leq 64\). Therefore, \(1 \leq r \leq 8\).
The volume of the solid can be found by integrating the function that defines the upper boundary (the cone) over the area defined by the ring. The integral setup will be:
\[
\int_0^{2\pi} \int_1^8 r \, r \, dr \, d\theta
\]
Explanation of solving this integral and calculating the volume can be detailed further on solving manually, ensuring to include bounds conversion, setup, and computation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5bc766f-fad7-4343-b342-499d4af4816f%2Fac4f5a64-90b9-41b7-b66f-4f533bb4b327%2Fwynh4qt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Use polar coordinates to find the volume of the given solid.
**Description of Solid:**
- The solid lies below the cone described by the equation:
\[
z = \sqrt{x^2 + y^2}
\]
- The solid is above the ring defined by the inequality:
\[
1 \leq x^2 + y^2 \leq 64
\]
**Instructions for Solution:**
To solve this problem, convert the Cartesian coordinates to polar coordinates. In polar coordinates, the relationships \(x = r\cos\theta\) and \(y = r\sin\theta\) are useful. The equation of the cone becomes \(z = r\), and the ring boundaries translate to \(1 \leq r^2 \leq 64\). Therefore, \(1 \leq r \leq 8\).
The volume of the solid can be found by integrating the function that defines the upper boundary (the cone) over the area defined by the ring. The integral setup will be:
\[
\int_0^{2\pi} \int_1^8 r \, r \, dr \, d\theta
\]
Explanation of solving this integral and calculating the volume can be detailed further on solving manually, ensuring to include bounds conversion, setup, and computation.
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