Triple integral volume problem. Please help me

In summary, the conversation discusses solving a triple integral problem to find the volume of a region in the first octant bounded by a plane and a surface. The conversation includes solving for the limits of integration and converting the integral to cylindrical coordinates. The final result is 12pi.
  • #1
VinnyCee
489
0
Ok I hve a triple integral problem for find the area of the following:

cylinder: [tex]x^2 + y^2 = 4[/tex]

plane: [tex]z = 0[/tex]

plane: [tex]x + z = 3[/tex]

It is a cylinder cut at the xy-plane and by the last plane. It looks like a circular wedge standing straight up from the xy-plane.

I just can't figure out the limits of integration, I know the last variable integrated must have limits that are constants. I just don't know what to do! Help me please! :tongue2:
 
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  • #2
Here is what I have so far

By solving the equation [tex]x + z = 3[/tex] for Z, [tex]z = 3 - x[/tex] I can get the z-limits to be from [tex]0[/tex] to [tex]3 - x[/tex], right?

I then solve [tex]x^2 + y^2 = 4[/tex] out to be [tex]y = \pm \sqrt{4 - x^2}[/tex], and those are the y-limits?

After that, the x-limits are from [tex]-2[/tex] to [tex]2[/tex], right?

Here is that in integral form: [tex]\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \int_{0}^{3 - x} dz dy dx = 12\pi[/tex]

Am I on the right track here?
 
  • #3
VinnyCee said:
By solving the equation [tex]x + z = 3[/tex] for Z, [tex]z = 3 - x[/tex] I can get the z-limits to be from [tex]0[/tex] to [tex]3 - x[/tex], right?

I then solve [tex]x^2 + y^2 = 4[/tex] out to be [tex]y = \pm \sqrt{4 - x^2}[/tex], and those are the y-limits?

After that, the x-limits are from [tex]-2[/tex] to [tex]2[/tex], right?

Here is that in integral form: [tex]\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \int_{0}^{3 - x} dz dy dx = 12\pi[/tex]

Am I on the right track here?

It looks perfect. :smile:

If you use cylindrical coordinates: r, phi, z, the integration becomes easier. In these coordinates

[tex] x=\cos\phi [/tex]
[tex] y= \sin \phi [/tex]
[tex] z=z [/tex],

the volume element is

[tex] dV= r d \phi dr dz [/tex]

and your boundaries are

[tex] 0 \le z \le 3-x = 3 - r \ cos\phi [/tex]

[tex] 0 \le \phi \le 2\pi [/tex]

[tex] 0 \le r \le 2 [/tex]

So you have to calculate the integral

[tex]\int_{0}^{2} \int_{0}^{2\pi}} \int_{0}^{3 - r \cos(\phi)}r dz d\phi dr [/tex]
 
  • #4
Here is the triple integral solved step-by-step

[tex]V = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \int_{0}^{3 - x} dz\; dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \left[ z \right]_{0}^{3 - x} dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \left[(3 - x) - (0)\right] dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} (3 - x) dy\; dx[/tex]

[tex]V = \int_{-2}^{2} \left[3y - xy\right]_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} dx = \int_{-2}^{2} \biggl\{\left[3\left(\sqrt{4 - x^2}\right) - x\left(\sqrt{4 - x^2}\right)\right] - \left[3\left(-\sqrt{4 - x^2}\right) - x\left(-\sqrt{4 - x^2}\right)\right]\biggl\} dx[/tex]

[tex]V = \int_{-2}^{2} \left(6 \sqrt{4 - x^2} - 2x \sqrt{4 - x^2}\right) dx = 6 \int_{-2}^{2} \left(4 - x^2\right)^{\frac{1}{2}}\; dx + \int_{-2}^{2} -2x \sqrt{4 - x^2}\; dx[/tex]

[tex]u = 4 - x^2[/tex] and [tex]du = -2x dx[/tex] for that second integral.

[tex]V = 6\left[\frac{2}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right)\right]_{-2}^{2} +\; 2\left[\frac{2}{3}\left(4 - x^2\right)^{\frac{3}{2}}\right]_{-2}^{2}[/tex]

[tex]V = 6\biggl\{\left{\sqrt{4 - 4} + 2 \sin^{-1}\left(-1\right)\right] - \left[\sqrt{4 - 4} + 2\sin^{-1}\left(1\right)\right]\biggl\} + 2\biggl\{\left[\frac{2}{3} \left(4 - 2^2\right)\right] - \left[\frac{2}{3} \left(4-(-2)^2\right)\right]\biggl\}[/tex]

[tex]V = 6\left(2\pi\right) + 2\left(0\right) = 12\pi[/tex]

Does this check out? Should I post a graph?
 
Last edited:
  • #5
Ok, a new triple integral

Did I do this new Problem correctly? :bugeye: :

Find volume of the region in the first octant bounded by the plane [tex]y = 1 - x[/tex] and the surface [tex]z = \cos\left(\frac{\pi\;x}{2}\right)[/tex]. And [tex]0\;\le x \;\le1[/tex].

Integral:

[tex]\int_{0}^{1} \int_{0}}^{1 - x} \int_{0}^{\cos\bigl\(\frac{\pi\;x}{2}}\bigl\)}\;dz\;dy\;dx = \frac{4}{\pi^2}[/tex]
 
  • #6
Anyone double check the last calculation yet?
 
  • #7
VinnyCee said:
Anyone double check the last calculation yet?

It looks all right.

ehild
 
  • #8
That is indeed correct

marlon
 
  • #9
Many thanks

Thank you ehild and marlon. I am never sure of some of these calc problems!
 

1. What is a triple integral volume problem?

A triple integral volume problem is a type of mathematical problem that involves calculating the volume of a three-dimensional object using triple integrals. It is commonly used in calculus and physics to solve problems related to finding the volume of complex shapes or regions.

2. How do I solve a triple integral volume problem?

To solve a triple integral volume problem, you need to set up and evaluate a triple integral based on the given boundaries and function. This involves breaking down the object into smaller, simpler shapes and using the properties of triple integrals to calculate the volume. It is a complex process that requires a good understanding of calculus and mathematical concepts.

3. What are the applications of triple integral volume problems?

Triple integral volume problems have various real-world applications, particularly in physics and engineering. They can be used to calculate the volume of irregularly shaped objects, such as a water tank, or to find the mass of a three-dimensional object with varying density. They are also used in computer graphics to render three-dimensional images and animations.

4. What are the challenges of solving triple integral volume problems?

Solving triple integral volume problems can be challenging because it requires a strong understanding of three-dimensional geometry and calculus. The process can also be time-consuming and prone to errors, especially when dealing with complex boundaries and functions. It is crucial to have a good grasp of the concepts and to double-check your calculations for accuracy.

5. Are there any tips for solving triple integral volume problems?

Some tips for solving triple integral volume problems include visualizing the three-dimensional object and breaking it down into simpler shapes, using symmetry to simplify the integral, and carefully setting up the boundaries and function. It is also helpful to practice with various examples and seek help from a tutor or online resources if you are struggling with the concept.

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