Answer to Question #159415 in Calculus for alex

Question

a)    Find the work done by the force field F on a particle that moves along the curve C. F(x, y) = (x² + xy)i + (y – x² y)j               C : x = t, y = 1/t (1 ≤ t ≤ 3)

b)    Use Green’s Theorem to evaluate the integral, assume that the curve C is oriented counterclockwise.  , where C is the triangle with vertices      (0, 0), (3, 3), and (0, 3).

c)     Evaluate the surface integral   , where σ is the part of the cone z =x² + y² that lies between the planes z = 1 and z = 2.

d)    Use the Divergence Theorem to find the flux of F across the surface σ with outward orientation.   F(x, y, z) = z³ i – x³ j + y³ k, where σ is the sphere x² + y² + z² = a².

Answer 1

a) $W=\intop_CF\cdot dr$

$r(t)=(t,1/t)$

$r'(t)=(1,-1/t^2)$

$F(t)=(t^2+1,\frac{1}{t}-t)$

$\displaystyle\intop_1^3(t^2+1,\frac{1}{t}-t)\cdot(1,-1/t^2)dt=\displaystyle\intop_1^3(t^2+1-\frac{1}{t^3}+\frac{1}{t})dt=$

$=(\frac{t^3}{3}+t+\frac{1}{2t^2}+lnt)|^3_1=9+1-\frac{1}{27}+\frac{1}{3}-1-1+1-1=8+\frac{8}{27}=\frac{224}{27}$

b)

$\oint_C(x^2+xy)dx+(y-x^2y)dy=\iint_D(Q_x-P_y)dA=$

$=\iint_D(-2xy-x)dA=\displaystyle\intop_0^3dy\displaystyle\intop_0^y(-2xy-x)dx=$

$=-\displaystyle\intop_0^3(y^3+\frac{y^2}{2})dy=-(\frac{y^4}{4}+\frac{y^3}{6})|^3_0=-(\frac{81}{4}+\frac{27}{6})=-\frac{297}{12}$

c) Parametri equations of the cone:

$x=zcost, y=zsint, z=z$ ($0\leq t\leq 2\pi, 1\leq z \leq 2$ )

$r(t,z)=zcosti+zsintj+zk$

$r_t\times r_z=\begin{vmatrix} i & j&k \\ -zsint & zcost&0 \\ cost&sint&1 \\ \end{vmatrix}=zcosti+zsintj-zk$

$|r_t\times r_z|=z\sqrt{2}$

$\iint_SdS=\iint_Q|r_t\times r_z|dA=\sqrt{2}\intop_0^{2\pi}dt\intop_1^2zdz=3\pi\sqrt{2}$

d)

$\iint_SF\cdot dS=\iiint_CdivFdV$

$divF=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}$

$divF=0$

$\iint_SF\cdot dS=0$