Integrating a Solid Enclosed by a Cylinder and Two Planes

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SUMMARY

The discussion focuses on calculating the volume of a solid enclosed by the cylinder defined by the equation x² + y² = 9 and the planes y + z = 5 and z = 1. The correct limits of integration for the triple integral are established as ∫_{-3}^3 ∫_{-√(9-x²)}^{√(9-x²)} ∫_{1}^{5-y} dz dy dx, leading to a volume of 36π. Participants emphasize the importance of switching to polar coordinates for simplification, with the integral expressed as ∫₀²ₚᵢ ∫₀³ (4 - r sin θ) r dr dθ.

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Familiarity with cylindrical coordinates
  • Knowledge of polar coordinate transformations
  • Basic integration techniques, including integration of polynomials
NEXT STEPS
  • Study the process of converting Cartesian coordinates to polar coordinates in integrals
  • Learn about finding limits of integration for triple integrals
  • Practice solving volume integrals involving cylindrical shapes
  • Explore advanced integration techniques, including trigonometric substitutions
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and multivariable integration, as well as educators seeking to enhance their teaching methods for volume calculations in three-dimensional spaces.

mateomy
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The solid enclosed by the cylinder x^2 + y^2 = 9 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...

<br /> \iiint dV<br />

<br /> \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{1}^{5-y} dzdydx<br />

<br /> \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (4-y) dydx<br />

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...

<br /> \int_0^{2\pi} \int_0^1 (4-rsin\theta)rdrd\theta<br />

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 x^2 + y^2 = 9 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...

<br /> \iiint dV<br />

<br /> \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{1}^{5-y} dzdydx<br />

<br /> \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (4-y) dydx<br />

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 \displaystyle \int(4-y)\,dy\,?
 
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 36\pi, 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 \displaystyle \int(4-y)\,dy\,?

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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