Volume Bounded by Cylinder and Plane

In summary: I'm getting close, just need to figure out the last bit.The integral will be over the region (-y,y) . So the final equation will be z=-y/2+y.
  • #1
MichaelT
25
0
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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  • #2
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:
 
  • #3
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?
 
  • #4
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:)
 
  • #5
Have you studied polar and cylindrical coordinates? Since this is a cylinder, cylindrical coordinates should simplify the calculations.
 
  • #6
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:
 
  • #7
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!
 
Last edited:
  • #8
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)
 
  • #9
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

[tex]\int\int(4-y[/tex]2)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...

[tex]\int[/tex](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:
 

What is the formula for finding the volume bounded by a cylinder and a plane?

The formula for finding the volume bounded by a cylinder and a plane is V = πr2h, where r is the radius of the base of the cylinder and h is the height of the cylinder.

How do you determine the height of the cylinder when given the volume and radius?

To determine the height of the cylinder when given the volume and radius, use the formula h = V / (πr2). Plug in the given values for the volume and radius to solve for the height.

What happens if the plane intersects the cylinder at an angle?

If the plane intersects the cylinder at an angle, the resulting shape will be a curved truncated cone. The formula for finding the volume in this case is more complicated and involves calculus.

Can the volume bounded by a cylinder and plane ever be negative?

No, the volume bounded by a cylinder and plane cannot be negative. Volume is always a positive quantity, representing the amount of space enclosed by a shape.

How does the volume bounded by a cylinder and plane change if the radius of the cylinder is doubled?

If the radius of the cylinder is doubled, the volume bounded by the cylinder and plane will also double. This is because volume is directly proportional to the radius squared in the formula V = πr2h.

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