Estimate the integral of g over the cube

[mathmari]

Hey!

I have the following exercise:
Integrate the $g=xyz$ over the cube that is on the first half-quadrant and it is bounded from the levels $x=1, y=1, z=1$.

Having the following formula:
$ \int \int_A{g(x,y,z)dS}= \int \int_D {g(x,y,z(x,y)) \sqrt{1+z_x^2+z_y^2}dxdy}$

do I have to take $z(x,y)=1$?
Then the integral is $ \int \int_A{xyzdS}= \int \int_D {xy \sqrt{1}dxdy}=\int \int_D {xy dxdy}$

Since the cube is on the first half-quadrant, $x \geq 0, y \geq 0$

So $ \int \int_A{xyzdS}= \int_0^1 \int_0^1 {xydxdy}=\frac{1}{4}$.

Do I have to do that also for taking $x(y,z)=1$ and then $y(x,z)=1$, and then add the results?

 

[I like Serena]

Hey!

I have the following exercise:
Integrate the $g=xyz$ over the cube that is on the first half-quadrant and it is bounded from the levels $x=1, y=1, z=1$.

Having the following formula:
$ \int \int_A{g(x,y,z)dS}= \int \int_D {g(x,y,z(x,y)) \sqrt{1+z_x^2+z_y^2}dxdy}$

do I have to take $z(x,y)=1$?
Then the integral is $ \int \int_A{xyzdS}= \int \int_D {xy \sqrt{1}dxdy}=\int \int_D {xy dxdy}$

Since the cube is on the first half-quadrant, $x \geq 0, y \geq 0$

So $ \int \int_A{xyzdS}= \int_0^1 \int_0^1 {xydxdy}=\frac{1}{4}$.

Do I have to do that also for taking $x(y,z)=1$ and then $y(x,z)=1$, and then add the results?

Hai!

Yep, yep, and yep.

Actually, you need to integrate over all 6 sides of the cube, but lucky for you, 3 of those sides will integrate 0 resulting in 0 so you can ignore them.

 

[mathmari]

Hai!

Yep, yep, and yep.

Actually, you need to integrate over all 6 sides of the cube, but lucky for you, 3 of those sides will integrate 0 resulting in 0 so you can ignore them.

To integrate over the 3 slides resulting in 0 do we take $x(y,z)=y(x,z)=z(x,y)=0$?

 

[I like Serena]

To integrate over the 3 slides resulting in 0 do we take $x(y,z)=y(x,z)=z(x,y)=0$?

Yep.
Btw, I am assuming you're supposed to integrate over the surface of the cube as opposed to the volume of the cube.

 

[mathmari]

Yep.
Btw, I am assuming you're supposed to integrate over the surface of the cube as opposed to the volume of the cube.

By the way I solved it did I calculate the volume of the cube?

 

[I like Serena]

By the way I solved it did I calculate the volume of the cube?

Nope. You integrated over the surface.
To integrate over the volume of the cube, you would need a triple integral:
$$\int_0^1\int_0^1\int_0^1 xyz\ dx dy dz$$

Edit: I'm just wondering if this is a setup for Gauss's theorem.

 

[mathmari]

Nope. You integrated over the surface.
To integrate over the volume of the cube, you would need a triple integral:
$$\int_0^1\int_0^1\int_0^1 xyz\ dx dy dz$$

Edit: I'm just wondering if this is a setup for Gauss's theorem.

Aha! Ok!
Thank you for your answer!

 

[HallsofIvy]

And, of course,
\int_0^1\int_0^1\int_0^1 xyz dxdydz= \int_0^1 xdx\int_0^1 ydy\int_0^1 zdz

 

[mathmari]

And, of course,
\int_0^1\int_0^1\int_0^1 xyz dxdydz= \int_0^1 xdx\int_0^1 ydy\int_0^1 zdz

Ok! Thanks a lot!