Integrating a Solid Enclosed by a Cylinder and Two Planes

In summary, The conversation is about finding the limits of integration for the solid enclosed by the cylinder x^2 + y^2 = 9 and the planes y + z = 5 and z=1. The conversation also includes a discussion about changing coordinates to find the limits and using trigonometric substitutions. The final answer for the integral is 36\pi, with a suggestion to use general advice for finding limits as a weak point.
  • #1
mateomy
307
0
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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  • #2
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]
 
  • #3
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.
 
  • #4
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.
 
  • #5
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)
 

FAQ: Integrating a Solid Enclosed by a Cylinder and Two Planes

What is the concept of volume via triple integrals?

The concept of volume via triple integrals is a mathematical method used to calculate the volume of a three-dimensional shape or region. It involves breaking down the shape into small parts and calculating the volume of each part using triple integrals. These individual volumes are then added together to get the total volume of the shape.

What are the applications of volume via triple integrals?

Volume via triple integrals is used in various fields such as physics, engineering, and mathematics. It is particularly useful in calculating the volume of objects with irregular shapes, such as fluid flow in a pipe, or the volume of a three-dimensional graph.

How is volume via triple integrals different from other methods of calculating volume?

Volume via triple integrals is a more advanced and versatile method compared to other methods such as calculating using geometric formulas or using cross-sectional area. It allows for the calculation of volume for complex and irregular shapes, whereas other methods may only work for simple shapes.

What are the steps involved in finding volume via triple integrals?

The steps for finding volume via triple integrals include:
1. Defining the limits of integration for the three variables (x, y, and z) based on the shape or region to be calculated.
2. Setting up the integral for the volume formula, which involves using the appropriate integrand and the limits of integration.
3. Evaluating the triple integral using the fundamental theorem of calculus.
4. Adding up the individual volumes to get the total volume of the shape or region.

Are there any limitations to using volume via triple integrals?

While volume via triple integrals is a powerful method for calculating volume, it does have some limitations. It can be quite complex and time-consuming to set up and evaluate the integrals, especially for shapes with intricate boundaries. Additionally, it may not be suitable for objects with varying densities or objects with changing volumes over time.

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