How Do I Set Up a Double Integral for a Cylinder's Volume in Polar Coordinates?

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Homework Help Overview

The discussion revolves around setting up a double integral in polar coordinates to find the volume of a cylinder defined by the region x² + y² ≤ 9 and 0 ≤ z ≤ 4. Participants are exploring how to incorporate the height of the cylinder into the integral setup.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up the double integral as ∫₀²π ∫₀³ 4r dr dθ, indicating a focus on the volume calculation. Some participants suggest variations in the integral limits and the inclusion of height in different ways, questioning how to properly express the volume in the context of the cylinder.

Discussion Status

There is an ongoing exploration of how to correctly set up the integral for the volume of the cylinder. Some participants have provided guidance on the structure of the integral, while others are still clarifying their understanding of the setup. Multiple interpretations of the integral limits and functions are being discussed.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may impose specific requirements on how the integral should be structured. There is a focus on ensuring that the height of the cylinder is appropriately represented in the integral.

whynot314
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Homework Statement


I want to convert this into polar and use double integral to find the volume of the solid in this region. I just need help setting this up
region
Q: x^2+y^2≤9, 0≤z≤4
I know this is a cylinder with a height of 4.
I am just having trouble incorporating this height into the integral.

The Attempt at a Solution


∫_0^2π▒〖∫_0^3▒4 r〗 drd(theta)
this is currently what I have
 
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"integral from 0 to 2pi" then integral 0 to 3. then 4 rdrdθ
 
This is correct.
There is a cylinder with height 4. When using a double integral to find the volume of a solid object, you can set it up with the "Top - Bottom" as the function to integrate. This can also be done by adding in a third integral and integrating 1.
\int_0^{2\pi} \int_0^3 \int_0^4 (1)dV, where dV is rdzdrd\theta.
=\int_0^{2\pi} \int_0^3 (4) (r)drd\theta
You can also check this by using the formula for the volume of a cylinder which is \pi r^2h
 
thank you
 
\int^\3_\0
 

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