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- #1

- Apr 14, 2013

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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?