Volume Bounded by Cylinder and Plane

MichaelT
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We need to find the volume of the solid bounded by the cylinder with the equation
z^2 + y^2 = 4 and the plane x + y = 2, in the first octant (x,y,z all positive).

Firstly, I am trying to visualize the graphs. From what I can tell, the cylinder is centered around the x-axis and has a radius of 2. It seems move along the x-axis infinitely. The plane intersects the cylinder at y=2, z=0. Visualizing this is tough and I haven't been able to find a graphing program that is sufficient.

Now, we have performed double integrals of solids over general two dimensional regions. I just cannot even think of where to start with this problem. Any help setting me in the right direction would be extremely helpful!
 
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MichaelT said:
We need to find the volume of the solid bounded by the cylinder with the equation
z^2 + y^2 = 4 and the plane x + y = 2, in the first octant (x,y,z all positive).

Hi MichaelT! :smile:

In a case like this, use either vertical or horizontal slices …

I suggest vertical, perpendicular to the x-axis …

what is the boundary of each slice? :smile:
 
Ok this is how I went about this. I used a double integral of the cylinder over the plane. I solved the equation of the cylinder for z, and that was the integrand. This was integrated over [0,2] X [0, 2-y] , dx dy.

After some lengthy calculations (and a very helpful table of integrals :approve:) I found the volume of the solid bounded by the cylinder and the plane to be 2pi - 8/3

I still need to check this over. I do have one question though, something I am not sure I did correctly. The integral of (4-y2)1/2dx = x(4-y2)1/2 Is this correct?
 
MichaelT said:
The integral of (4-y2)1/2dx = x(4-y2)1/2 Is this correct?

Yes, that's right. :smile:

(but if you want us to check the rest, you'll need to type it out in full, so we can see it! :wink:)
 
Have you studied polar and cylindrical coordinates? Since this is a cylinder, cylindrical coordinates should simplify the calculations.
 
I'm pretty sure I got it right, since a classmate of mine solved for the volume using a triple integral and we both got the same answer. So I don't think I need it checked, but if anyone wants to see the whole thing I would be happy to post it.

As to polar and cylindrical coordinates...We have studied them but not to the extent that I would feel confident using them for this problem. Thanks for the tip though, I am always interested in other approaches to problems :biggrin:
 
hey i would like to see how to do this problem ..can someone post a worked out solution or give me step by step directions how to complete it so i can try myself. I'm having trouble setting the problem up as a triple integral. After this point i should be good.

Thanks!
 
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Hey negat1ve!

I haven't done this problem with a triple integral, but I did do it with a double integral. If you would like to use that approach (a bit easier, as long as you have an integration table handy :wink:)...Think of the plane x + y = 2 in two dimensions. What shape does this form in the first quadrant? (since we are dealing with all x, y, and z positive)
 
yeah I am graphing the equation it looks like a diamond or a parallelogram
 
  • #10
Hmmm well in 3 dimensions the plane x + y = 2 is a plane. But if you want to look at it in 2 dimensions, just plot some points in the first quadrant of the xy coordinate system. i.e when y=0, x=2.

If you take the points and connect them, a common geometric shape will be bounded between the x axis, y axis, and the line. This will help you find the bounds of integration.
 
  • #11
hey ok so it makes a right triangle in the first quad right
 
  • #12
Sure does! Now think about the bounds of this triangle. Have you studied x-simple and y-simple regions? You can think of the triangle as both y-simple and x-simple because it ends in either a straight line or a point in both the x and y directions. So it is up to you to choose. In one direction the bounds will be integers, and in the other it will be a function.

So what will the integrand be? It will help you determine which way to think about the triangle.
 
  • #13
So i want to use

ok so would this be my integrand after solving the cylinder for z

\int\int(4-y2)1/2dxdy


and i would integrate over the reigion R where

R= [0,2-y] x [0,2]?

I however see that you posted your region R rectangle opposite of what i stated here.

Can you explain this?
 
  • #14
So basically i come down to doing this nasty integral...

\int(4-y2)1/2 2-y dy

Am i on the right track...

What is this integration by parts?
 
  • #15
integration by parts

negat1ve said:
What is this integration by parts?

Who mentioned integration by parts? :confused:

Integration by parts is the inverse version of the product rule in differentiation …

see http://en.wikipedia.org/wiki/Integration_by_parts :smile:
 
  • #16
Well am I at least proceeding correctly? Should I have come up to this integral to even have to calculate. If so how can i solve this integral?

If not what did i do wrong can you please explain...

I never said it should be done using integration by parts... i was just saying it looked like it might be able to be solved that way... It can't be done using a u substitution. Is this where a table of integrals would come in handy:confused:

Please I am serious about wanting to know how to solve this problem...
 
  • #17
negat1ve said:
Well am I at least proceeding correctly? Should I have come up to this integral to even have to calculate. If so how can i solve this integral?

I'm not following your formula …

take either horizotal or vertical slices …

I suggest vertical slices of thickness dx, each of which should then be a triangle and an arc-sector of a circle. :smile:
I never said it should be done using integration by parts...


I never said you said it should! … :biggrin:
 
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