Use Green's Theorem to calculate the work done by the force 𝐅 on a particle that is moving count

Use Green's Theorem to calculate the work done by the force...

Key Concepts

Green's Theorem

Green's Theorem is a fundamental result in vector calculus that relates a line integral around a simple, closed curve in the plane to a double integral over the region enclosed by the curve. It converts the circulation (or work) computed along the boundary into an area integral of the partial derivatives (specifically, the difference between the partial derivative of one component of the vector field with respect to one variable and that of the other component with respect to the other variable), making many complex line integrals easier to compute.

Line Integrals in Vector Fields

Line integrals in vector fields measure the work done by a force field along a curve or the circulation of the field around a closed path. When calculating the work done by a force, the line integral accumulates the contributions of the force's components along the direction of displacement, providing a central tool for linking vector fields to physical quantities like work and flux.

Double Integrals

Double integrals extend the concept of integration to two-dimensional regions, allowing for the summation of a function’s values over an area. In the context of Green's Theorem, the original line integral is transformed into a double integral over the region bounded by the curve, which often simplifies the calculation by converting a potentially complicated boundary integral into a more straightforward area-based computation.

Coordinate Transformation (Polar Coordinates)

Changing coordinate systems, such as to polar coordinates, is a useful technique when the region of integration exhibits circular or radial symmetry. By expressing the variables in terms of radius and angle, the area element and the limits of integration can be simplified, which makes evaluating the double integral more tractable when the domain is defined by curves naturally expressed in polar form.

Transcript

00:01 In this question, we need to calculate the work done in moving a particle along the curve given as c, r equals to 2 cosine theta.

00:13 And if the expression for force, that is, it is actually a vector, is given as the function of x and y as equal to e to the power x minus 3 by i cap plus e to the power y plus 6 y cap.

00:29 Now we know that the work done can be calculated as that is integral of f dot d s that is displacement.

00:43 So in the vector form this will be calculated as the vector along x direction to be integrated with respect to x and vector along y direction to be integrated with respect to y.

00:56 Therefore the work done can be calculated as work done equal to the integral over the curve that is c f dot d x will be equal to f dot d s sorry f dot ds will be equal to that is e to the power x minus 3y dot d x plus e to the power y plus 6y dot now using green's theorem actually we need to use green's theorem to calculate the work in this question we will get this integral we know that green's theorem state that the integral of a function that is m.

01:52 Dx over the curve c the integral of m.

01:56 Dx plus n .diy can also be evaluated as double integral over the region are the integral of partial derivative of n there is del n over del x minus del m over del y partial derivative of m with respect to y dot d a where a is the area enclosed by the curve c so from here the given expression for the work we can write the value of m that is expression for m will be e to the power x minus 3y and similarly expression for n will be e to the power y minus so it is 6x not 6y so right here 6x not 6y here again 6x now we need to calculate the partial derivatives of m with respect to y and n with respect to x we will get del m over del y will be equal to e to each the power x will be differentiated to 0 as we are calculating with respect to y so we will get del m over del y as minus 3 and del n over del x will be equal to e to the power y will be differentiated to 0 and 6x will be differentiated to 6 so we got the values of partial derivatives that is del n over del x is 6 and del m over del y is minus 3 and we know the given curve is c is given as that is the closed path c is given as r equals to 2 cosine theta and we can multiply r both side we will get r square equals to 2 r cosine theta or we can write from here that x squared plus y square will be equal to 2x as r square will be replaced by x square plus y square equal to to r cos theta will be replaced by two x so here r cosine theta will be replaced by x so it becomes 2x and on simplifying this we will get the equation of this circle as x square minus 2x plus y square equals to 0 and we can write it in a standard form as that is x square minus 2x plus 1 plus y square equals to 1 and this can be written as x minus 1 whole square plus y square equals to 1 and in this circle the circle is having radius 1 and center at 1 .0 so radius of this circle is 1 and center of the circle is at that is 1 .0 so we got the coordinates of center and radius of the circle we know that the work done we already calculated as f dot d r can be written as the work done will be equal to that is integral over the curve c f dot d r or that we already written as work done will be equal to using green's theorem double integral over the integral region r of del n over del x minus del m over del y dot d a we already calculated del n over del x and del m over del y so we will get this work done as double integral and for the reason r and del n over del x was 6 and del m over del y was minus 3 so this will become 6 minus minus 3 that is plus 3 dot d a so from here we can simplify this we will get this work done will be equal to nine times the double integral for the region r .d .a...

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