Multivariable Calculus - Integration Assignment 1#

[ConnorM]

Homework Statement

Evaluate the integral,

\iiint_E z dzdydz

Where E is bounded by,

y = 0
z = 0
x + y = 2
y^2 + z^2 = 1

in the first octant.

Homework Equations

Rearranging y^2 + z^2 = 1 it terms of z,
z = \sqrt{1-y^2}

The Attempt at a Solution

From the given equations I determined that my bounds were,

1 \leq x \leq 2
0 \leq y \leq 1
0 \leq z \leq \sqrt{1-y^2}

I found these bounds by first looking at z = \sqrt{1-y^2} and seeing that y must be between 0 and 1 since we are working in the first octant, also z must be between 0 and z = \sqrt{1-y^2}. Then I moved on to x + y = 2, since y can only be between 0 and 1 the only way for the equation x + y = 2 to be true is if x is between 1 and 2.

\int_1^2 \int_0^{2-x} \int_0^\sqrt{1-y^2} z dzdydz

After integrating I found my answer to be 1/3. Can anyone let me know if I've made a mistake anywhere or if I have done this correctly?

 

[DarthRoni]

I'm doing the same assignment. I also got 1/3

EDIT: I'm not so sure about that answer anymore

 

[HallsofIvy]

Then I moved on to x+y=2, since y can only be between 0 and 1 the only way for the equation x+y=2 to be true is if x is between 1 and 2.

I get that x must be between 0 and 1. Otherwise it won't be under the cylinder y^2+ z^2= 1.
\int_0^1\int_0^{2- x}\int_0^{\sqrt{1- y^2}} zdzdydx

 

[STEMucator]

Indeed, if you project the three planes and cylinder into the positive octant, you can observe that $0 \leq x \leq 1$. You can check this out in the images I attached to help visualize. Try to see how the planes cut the cylinder, this is what let's you determine your limits most of the time.