# Thread: some questions for the final

1. ## some questions for the final

True or False. (why)

1. There is a vector field F in $\displaystyle R^3$ such that curl(F)= y i + x j +z k. why?

2. Let D1 be an oriented surface and D2 be the same surface with opposite orientation. Then, for every integrable function f, we have $\displaystyle \int\int_{\omega{1}}f dS =-\int\int_{\omega{2}}f dS$ why?

My thought: if f is scalar function, then this statement would be false?? Does this surface has to be a closed surface?

3. For any smooth vector field F on a torus $\displaystyle \omega \subset R^3 , \int\int_{\omega} curl(F)\cdot dS$ is zero
My thought: use Stoke's theorem? omega has to be no boundary?

4. Let$\displaystyle F(x, y, z)=2x \imath - siny \jmath + z cosy \kappa$ and $\displaystyle \sum_{R}$ be the sphere $\displaystyle x^2 +y^2 + z^2 = R^2$ oriented with outward-pointing normal. Then, the flux $\displaystyle \int\int_{\sum_{R}}F \cdot dS$ increases with the radius R.

2. Originally Posted by yzc717
True or False. (why)

1. There is a vector field F in $\displaystyle R^3$ such that curl(F)= y i + x j +z k. why?
false. for a vector field $\displaystyle F$ we must have $\displaystyle \text{div}(\text{cur}(F))=0$ but in here we have $\displaystyle \text{div}(\text{cur}(F))=1.$

2. Let D1 be an oriented surface and D2 be the same surface with opposite orientation. Then, for every integrable function f, we have $\displaystyle \int\int_{\omega{1}}f dS =-\int\int_{\omega{2}}f dS$ why?
true. properties of surface integral. see your lecture notes!

3. For any smooth vector field F on a torus $\displaystyle \omega \subset R^3 , \int\int_{\omega} curl(F)\cdot dS$ is zero
true. the divergence theorem + see my answer to part 1 of your problem.

4. Let$\displaystyle F(x, y, z)=2x \imath - siny \jmath + z cosy \kappa$ and $\displaystyle \sum_{R}$ be the sphere $\displaystyle x^2 +y^2 + z^2 = R^2$ oriented with outward-pointing normal. Then, the flux $\displaystyle \int\int_{\sum_{R}}F \cdot dS$ increases with the radius R.
true. by the divergence theorem the value of your surface integral is $\displaystyle \frac{8}{3}\pi R^3.$