A particle moves along line segments from the origin to the points (1, 0, 0), (1

Question

A particle moves along line segments from the origin to the points (1, 0, 0), (1, 4, 1), (0, 4, 1), and back to the origin under the influence of the force field F(x, y, z) = z^2(i) + 3xy(j) + 2y^2(k). Find the work done.

Explanation / Answer

ans is : 126 pie The curve would enclose a rectangular shaped region on the plane -y + 5z = 0. I got this plane by noting that it would contain the vectors u = (3,5,1) - (3,0,0) = (0,5,1) and v = (0,5,1) - (3,5,1) = (-3,0,0). The normal direction is given by u x v = (0, -3, 15). As the plane contains the origin, it has equation 0(x) - 3(y) + 15(z) = 0 ==> -y + 5z = 0. This is a level surface. Stoke's theorem says that ? F·dr = ?? curl(F)·n dS C.........S where S is any surface that has the contour C as a boundary and n is the outward normal to S. If you draw a sketch, it becomes clear that to have the curve traversed as stated the right hand rule says that the normal should have a positive component in the k direction. Letting g(x,y,z) = -y + 5z =0 denote the level surface, the normal and differential surface area are n = grad(g)/||grad(g)|| and dS = ||grad(g)|| dA so ndS = grad(g) dA. You can project the plane into the xy plane or the xz plane (but not into the yz plane since it is parallel to the x-axis) to get dA. I'll choose the xy plane. The region in the xy-plane that is mapped to our portion of the plane -y + 5z = 0 is the rectangle 0 = x = 3, 0 = y = 5. So n dS = (0, -1, 5) dydx and the limits are 0 = x = 3, 0 = y = 5. The curl of F is curl(F) = (10y, 2z, 4y). So curl(F)·ndS = (-2z + 20y) dydx. On the plane z = y/5 giving -2z + 20y = 98y/5, so the work is 3 5 ? ? (98/5) y dy dx = 735. 0 0 **** Update ****** I found an error in the differential, so the answer is off by a factor of 1/5. The integral should be 735/5 = 147. If integrating in x and y, then it should read dS = ||grad(g)||/|grad(g)·k| dy dx and if integrating in z and x it should be dS = ||grad(g)||/|grad(g)·j| dz dx. In any case, ||grad(g)|| = v(1² + 5²) = v(26), and |grad(g)·k| = 5 while |grad(g)·j| = 1. Integrating in x and y n = (0, -1, 5)/v(26) and dS = v(26)/5 dy dx So the work is 3 5 ? ? (98/5²) y dy dx = 147. 0 0 Alternatively integrating in z and x n = (0, -1, 5)/v(26) and dS = v(26) dz dx 3 1 ? ? 98 z dz dx = 147 0 0

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