cuboid with $z=0$ and $z=y$

Question

Compute $\int \int _S F \cdot n \hspace{2mm} dS$ where $$F(x,y,z)=(x-z\cos y, y-x^2+x\sin z+z^3, x+y+z)$$ and $r$ is the surface that bounds the solid between the planes $$x=0,\hspace{2mm}x=1,\hspace{2mm}y=0,\hspace{2mm}y=2,\hspace{2mm}z=0, \hspace{2mm} z=y$$

Are there typos in this? What is $r$? And I found it kind of weird that z=y. The divF=3 so do we just do (using divergence theorem) $$3\int \limits_{x=0}^1 \int \limits_{y=0}^2 \int \limits_{z=0}^y dzdydx$$ which gives $6$ right?

Answer 1

Right!

I wrote $$F = (x - zcos(y),y - {x^2} + xsin(z) + {z^3},x + y + z)$$ as a 2-form: $$\omega = (x - cos(y)z)dy \wedge dz + (y - {x^2} + xsin(z) + {z^3})dz \wedge dx + (x + y + z)dx \wedge dy$$

calculate exterior derivative:

$$\begin{gathered} d\omega = dx \wedge dy \wedge dz + dy \wedge dz \wedge dx + dz \wedge dx \wedge dy \hfill \\ \hfill \\ d\omega = 3\cdot dx \wedge dy \wedge dz \hfill \\ \end{gathered}$$

applied Stoke's theorem and got:

$$\begin{gathered} \int\limits_S \omega = \int\limits_V {d\omega } = \int\limits_V {3\cdot dx \wedge dy \wedge dz} = 3 \cdot \int\limits_V {dx \wedge dy \wedge dz} \hfill \\ \hfill \\ = 3 \cdot \int\limits_0^1 {\int\limits_0^2 {\int\limits_0^y {dz} } } \cdot dy \cdot dx = 3 \cdot \int\limits_0^1 {\int\limits_0^2 {y\cdot dy \cdot dx} } \hfill \\ \hfill \\ = 3 \cdot \int\limits_0^1 {\frac{{{2^2}}}{2}\cdot dx} = 3 \cdot 2 \cdot \int\limits_0^1 {dx} = 3 \cdot 2 \cdot 1 = 6 \hfill \\ \end{gathered}$$

the same!

Answer 2

Your calculations are correct. I suspect that $r$ was supposed to be $S$. There's nothing strange about having $z = y$ as one of the bounds -- it's a plane that makes a $45^{\circ}$ angle relative to the $xy$-plane.