Answer to Question #232009 in Calculus for Anuj

Question

Consider the surface S =  (x, y, z) ∈ R 3 | z = 3 − x 2 − y 2 ; z ≥ 2 . Assume that S is oriented upward and let C be the oriented boundary of S. (a) Sketch the surface S in R 3 . Also show the oriented curve C and the XY-projection of the surface S on your sketch. (2) (b) Let F (x, y, z) = (2y, 3z, 4y). Evaluate the flux integral Z Z S (curl F)

Answer 1

(a)

(b)

$\text{Given } Z=g(x,y)=3-x^2-y^2, z\ge2 \\ \text{for this problem } \\ \hspace{30pt} g_x=-2x \text{ and } g_y=-2y \\ \text{If follows that the normal vector is } \left\langle { - 2x, - 2y, - 1} \right\rangle \\ \implies \overrightarrow F \cdot \overrightarrow n = \left\langle {2y,3z,4y} \right\rangle \cdot \left\langle { - 2x, - 2y, - 1} \right\rangle =-4xy-6yz-4y \\ \implies \text{Flux }=\iint_R {\left( { - 4xy - 6yz - 4y} \right)dA} \\ \text{The region R is the disk of radius } \sqrt{2} \le r \le \sqrt{3}, \text{ and } 0 \le \theta \le 2\pi. \\\text{Substituting } x=r\cos\theta \text{ and } y=r\sin\theta, \\ \text{Flux }= \int_0^{2\pi } {\int_{\sqrt 2 }^{\sqrt 3 } {\left( { - 4{r^2}\sin \theta \cos \theta - 6r\sin \theta \left( {3 - {r^2}} \right) - 4r\sin \theta } \right)rdrd\theta } } \\ \hspace{25pt} = \int_0^{2\pi } {\int_{\sqrt 2 }^{\sqrt 3 } {\left( { - 4{r^3}\sin \theta \cos \theta - {r^2}\sin \theta \left( {22 - 6{r^2}} \right)} \right)drd\theta } } \\ \hspace{25pt} = \int_0^{2\pi } {\left[ { - {r^4}\sin \theta \cos \theta + \frac{{6{r^5}}}{5}\sin \theta - \frac{{22{r^3}}}{3}\sin \theta } \right]} _{\sqrt 2 }^{\sqrt 3 }d\theta \\ \hspace{25pt} = \int_0^{2\pi } {\frac{{ - 56\sqrt 3 }}{5}} \sin \theta - 5\sin \theta \cos \theta + \frac{{148}}{{15}}\sqrt 2 \sin \theta d\theta \\ \hspace{25pt} =\left[ { - \frac{{148\sqrt 2 }}{{15}}\cos \theta + \frac{{56\sqrt 3 }}{5}\cos \theta - \frac{5}{2}{{\sin }^2}\theta } \right]_0^{2\pi } \\ \hspace{25pt} = 0.$