![]() ![]() My integral is going in the clockwise direction instead of the counterclockwise, so I will place a negative in front of it. The double integral uses the curl of the. The line integral in question is the work done by the vector field. I'm left with -12cos^3t if I did my math correctly. First we will give Greens theorem in work form. Lots of stuff cancels! I wound up with -18costsint + 18cos^2t + 18costsint - 18cos^2t - 12cos^3t. My integral will take the dot product of F(r(t)) and r'(t). I found that F(r(t)) is 3costi + (-6sint+6cost)j - 6cos^2tk. I also found F(r(t)) by sticking the components of r(t) into those of F. As with the last section we will start with a two-dimensional curve C C with parameterization, x x(t) y y(t) a t b x x ( t) y y. ![]() In this section we want to look at line integrals with respect to x x and/or y y. I proceeded to find the derivative of r(t) and found that r'(t) = (-6sint + 6cost)i -3costj + 2costk. In the previous section we looked at line integrals with respect to arc length. Thus I believe my lower limit will be 3 pie / 2. It follows then that this also satisfies y = 3cost = 0 and x = 6cost + 6sint = -6. Thank you very much for the help! That really helps me out, and I appreciate it.įor the lower limit, I noticed that z = sint = -1 is satisfied by 3 pie / 2. I don't think that will be that difficult, especially if I can simplify some (for instance, I see I could simplify the i component of my hpothetical curve to 6(cost+sint)i.Īny assistance would be much appreciated! I know that once I get an r(t), I need to stick it into my function F, and then dot itself with the derivative of r(t). More than likely simple, but I'm stumped. In classical mechanics, line integral is used to compute the word performed on mass m moving in a gravitational field. Line integral workdone how to#That's where I come to an abrupt stop.because I need r(t) going from (6,0,2) to (-6,0,-2), and I'm not quite sure how to do this. Line integrals have several applications such as in electromagnetic, line integral is used to estimate the work done on a charged particle travelling along some curve in a force field defined by a vector field. I then stuck in the y and z components I had (3cost and 2 sint, respectively) and came up with this curve: (6cost+6sint)i + 3costj + 2sintk. In order to find a curve formed by the intersection, I used 3costj + 2sintk and then, for my i component, I noted that the plane equation gave me x = 2y + 3z. My main stumping point is coming up with the correct parameterization for r(t). I just need to set the integral up correctly, because I'm going to be letting the computer do the integration. I'm going to need to find the integral of F dr which is actually going to be the integral of Fr(t) dot with r'(t) dt. Find the work done on it by the vector field F(x,y,z) = -yi + xj + yzk. The particle moving along the curve goes from (6,0,2) to (-6,0,-2). All these things can be taken into account by defining work as an integral.A curve is formed by the intersection of y^2/9 + z^2/4 = 1 and the plane x-2y-3z = 0. ![]() The general definition of work done by a force must take into account the fact that the force may vary in both magnitude and direction, and that the path followed may also change in direction. The power of calculus can also be applied since the integral of the force over the distance range is equal to the area under the force curve:įor any function of x, the work may be calculated as the area under the curve by performing the integral For example, for the work done to stretch a spring, the area under the curve can be readily determined as the area of the triangle. In the more general case of a force which changes with distance, the work may still be calculated as the area under the curve. That relationship gives the area of the rectangle shown, where the force F is plotted as a function of distance. The basic work relationship W=Fx is a special case which applies only to constant force along a straight line. Work as an integral Work done by a variable force ![]()
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