Use a double integral in polar coordinates to find the volume V of the solid bounded by the graphs of the equations. √x² + y² V = Z = √x Z = 0 x² + y² = 4 NR X 2 √x² +1² X dr de =
Use a double integral in polar coordinates to find the volume V of the solid bounded by the graphs of the equations. √x² + y² V = Z = √x Z = 0 x² + y² = 4 NR X 2 √x² +1² X dr de =
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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![**Transcription for Educational Website**
**Use a double integral in polar coordinates to find the volume \( V \) of the solid bounded by the graphs of the equations.**
Equations:
\[ z = \sqrt{x^2 + y^2} \]
\[ z = 0 \]
\[ x^2 + y^2 = 4 \]
Integral for Volume:
\[ V = \int_0^{2\pi} \int_0^2 \sqrt{x^2 + y^2} \, dr \, d\theta \]
**Explanation:**
- The problem involves finding the volume of a solid using integration in polar coordinates.
- The equations given describe a surface, a plane, and a circle.
- The correct setup for the double integral is:
- The outer integral limits for \( \theta \) are from 0 to \( 2\pi \), representing a full rotation around the circle.
- The inner integral limits for \( r \) are from 0 to 2, covering the radius from the center to the edge of the circle in the \( xy \)-plane.
- The integral involves calculating over the region defined by the circle \( x^2 + y^2 = 4 \).
- \( \sqrt{x^2 + y^2} \) is the function to be integrated, corresponding to the height \( z \) of the solid at each point in the circular region.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa48f7eb-bce8-4ea0-bf8b-7482c9c56a7c%2Ffe089bbe-2c7b-475d-8d7c-cc4f74b8dbd4%2Ffu0294r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
**Use a double integral in polar coordinates to find the volume \( V \) of the solid bounded by the graphs of the equations.**
Equations:
\[ z = \sqrt{x^2 + y^2} \]
\[ z = 0 \]
\[ x^2 + y^2 = 4 \]
Integral for Volume:
\[ V = \int_0^{2\pi} \int_0^2 \sqrt{x^2 + y^2} \, dr \, d\theta \]
**Explanation:**
- The problem involves finding the volume of a solid using integration in polar coordinates.
- The equations given describe a surface, a plane, and a circle.
- The correct setup for the double integral is:
- The outer integral limits for \( \theta \) are from 0 to \( 2\pi \), representing a full rotation around the circle.
- The inner integral limits for \( r \) are from 0 to 2, covering the radius from the center to the edge of the circle in the \( xy \)-plane.
- The integral involves calculating over the region defined by the circle \( x^2 + y^2 = 4 \).
- \( \sqrt{x^2 + y^2} \) is the function to be integrated, corresponding to the height \( z \) of the solid at each point in the circular region.
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