Integrating a Solid Enclosed by a Cylinder and Two Planes

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Homework Help Overview

The discussion revolves around integrating a solid defined by the cylinder x² + y² = 9 and the planes y + z = 5 and z = 1. Participants are exploring the limits of integration and the potential need to switch to polar coordinates for the integration process.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss finding limits of integration and express uncertainty about switching to polar coordinates. Some are attempting to set up integrals in both Cartesian and polar forms, while others question the necessity of changing coordinates.

Discussion Status

The conversation is ongoing, with participants sharing their attempts at setting up integrals and discussing the challenges they face. Some guidance has been offered regarding the limits of integration, but there is no explicit consensus on the best approach yet.

Contextual Notes

Participants mention feeling intimidated by the limits of integration and the complexity of the integrals involved, indicating a need for further clarification on these aspects.

mateomy
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The solid enclosed by the cylinder [itex]x^2 + y^2 = 9[/itex] and the planes y + z = 5 and z=1.

The biggest part for me (usually) is just being able to find my limits of integration for these problems (any suggestions about that would also be greatly appreciated). I think I found the correct limits for this problem...

[tex] \iiint dV[/tex]

[tex] \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{1}^{5-y} dzdydx[/tex]

[tex] \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (4-y) dydx[/tex]

At this point I start to get lost. Should I switch it to polar coordinates? I tried to do that from the last step above and it came out wrong. Here's my first step into the polar coordinate switch...

[tex] \int_0^{2\pi} \int_0^1 (4-rsin\theta)rdrd\theta[/tex]

Does this look like I'm headed in the right direction? This chapter is completely confusing me.
 
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mateomy said:
The solid enclosed by the cylinder [itex]x^2 + y^2 = 9[/itex] and the planes y + z = 5 and z=1.

The biggest part for me (usually) is just being able to find my limits of integration for these problems (any suggestions about that would also be greatly appreciated). I think I found the correct limits for this problem...

[tex] \iiint dV[/tex]

[tex] \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{1}^{5-y} dzdydx[/tex]

[tex] \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (4-y) dydx[/tex]

At this point I start to get lost. Should I switch it to polar coordinates? I tried to do that from the last step above and it came out wrong.
...
Why change coordinates?

That's a basic integral.

What is [itex]\displaystyle \int(4-y)\,dy\,?[/itex]
 
I'm doing integral[0,2pi], integral[0,3], integral[1,5-rsintheta] of 1 dz r dr dtheta
Sorry for the mess. Don't know how to display the integral sign. I got 36 pi.
 
eurekameh said:
I'm doing integral[0,2pi], integral[0,3], integral[1,5-rsintheta] of 1 dz r dr dtheta
Sorry for the mess. Don't know how to display the integral sign. I got 36 pi.

Yeah, those limits make sense. The answer is [itex]36\pi[/itex], so you got it. My integrals started looking insane so I figured -rightly- that I was doing something wrong. Thanks.

Have any general advice for finding the limits? That seems to be my biggest weak-point.
 
SammyS said:
Why change coordinates?

That's a basic integral.

What is [itex]\displaystyle \int(4-y)\,dy\,?[/itex]

I know that basic integral, but the limits around it make it really intimidating because you'd have to end up using trig-subs. Right? (Thanks for the help, btw)
 

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