# Volume of a cylinder restriced by planes

• Jul 16th 2008, 04:17 PM
crabchef
Volume of a cylinder restriced by planes
Find the area of a cylinder x^2+y^2=9 , the planes y+z=1 and z=1

This is how I set it up: ∫ ∫ ∫ (x^2+y^2) dz dy dx

=> z bounded by 1 < z < 5-y

∫ ∫ ∫ (x^2+y^2) dz dy dx = ∫ ∫ (4-y)(x^2+y^2) dy dx

=> using x = r cos T, y = r sin T, I get:

∫ ∫ (4-rsin T)(r^2)(r) dr dT where 0 < r < 3, 0 < T < 2pi.

I try to calculate this and I get the wrong answer. Did I set this up incorrectly? I was hoping it was just a computational error, but I have double checked it and still get the wrong answer, which is supposed to be 36pi.

Thanks!
• Jul 16th 2008, 04:58 PM
crabchef
Thanks for the help, but I have another question ><

Doesn't the fact that the cylinder is bounded by z=5-y matter? I don't see how that is built into the integral you wrote up there (Thinking)
• Jul 16th 2008, 05:04 PM
mr fantastic
Quote:

Originally Posted by crabchef
Find the area of a cylinder x^2+y^2=9 , the planes y+z=1 and z=1

This is how I set it up: ∫ ∫ ∫ (x^2+y^2) dz dy dx

=> z bounded by 1 < z < 5-y

∫ ∫ ∫ (x^2+y^2) dz dy dx = ∫ ∫ (4-y)(x^2+y^2) dy dx

=> using x = r cos T, y = r sin T, I get:

∫ ∫ (4-rsin T)(r^2)(r) dr dT where 0 < r < 3, 0 < T < 2pi.

I try to calculate this and I get the wrong answer. Did I set this up incorrectly? I was hoping it was just a computational error, but I have double checked it and still get the wrong answer, which is supposed to be 36pi.

Thanks!

The correct set up for the integrals is:

$\displaystyle V = \int \int_{R_{xy}} \int_{z = 1 - y}^{z=1} dx\, dy, dz$

where $\displaystyle R_{xy}$ is the upper half of the circle $\displaystyle x^2 + y^2 = 9$ in the xy-plane.

The surfaces defining the volume get built into the integral terminal, NOT the integrand (which is the mistake you made).
• Jul 16th 2008, 05:06 PM
mr fantastic
Quote:

Originally Posted by crabchef
Thanks for the help, but I have another question ><

Doesn't the fact that the cylinder is bounded by z=5-y matter? I don't see how that is built into the integral you wrote up there (Thinking)

You've gave the plane y+z=1 in your original question. So which is it: y + z = 1 or y + z = 5?
• Jul 16th 2008, 08:35 PM
crabchef
ah sorry! it was in fact y + z =5
• Jul 16th 2008, 11:01 PM
mr fantastic
Quote:

Originally Posted by crabchef
ah sorry! it was in fact y + z =5

Draw a large and well labelled diagram (I found a side view in the zy-plane extremely helpful) and you easily see that V = V1 + V2 where:

V1 is the volume of a cylinder of height 1 and radius 3: $\displaystyle V_1 = 9 \pi$.

V2 is half the volume of a cylinder of height 6 and radius 3: $\displaystyle V_2 = \frac{1}{2} \pi (3)^2 (6) = 27 \pi$.

Alternatively:

$\displaystyle V = \int \int_{R_{xy}}(5 - y) - 1 \, dx \, dy = \int \int_{R_{xy}}4 - y \, dx \, dy$

where $\displaystyle R_{xy}$ is the region of the xy-plane defined by the circle $\displaystyle x^2 + y^2 = 9$.
• Jul 16th 2008, 11:45 PM
crabchef
thanks for the help. It helps a bit, but I'm still having a lot of trouble in this class (just your basic multi var course)! ugh it's frustrating...