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.