Example 1 : Show that the vector field is conservative. Find its potential and also the work done in moving a particle from (1, 0) to (2, 1) along some curve. Solution : Vector field is conservative if around any closed curve is always zero By stokes theorem, So, for conservative field So, the potential …
Example 1 : Evaluate where C is any path from (0, 0, 1) to . Solution : Here, Since, curl So, is conservative vector field. So, the given line integral is independent of path. Let us find the potential corresponding to . So, …(1) …(2) …(3) Adding (1), (2), (3) and adding the common terms …
Let be a vector point function defined and continuous in region R of space. Let A & B are two points in R. Let and are arbitrary paths joining A & B. A vector field is said to be conservative if from A to B is independent of the paths. In this case the value …
Example 1 : Verify Stoke’s theorem for where S is the upper half surface of the sphere and C is its boundary. Solution : The surface S is the part of sphere above xy plane bounded by curve C, as shown in fig. On curve C, So, Now, let us evaluate surface integral Now, consider …
Example 1 : Verify Stoke’s theorem for taken round the rectangle bounded by Solution : C is a piecewise smooth curve consisting of y = 0, x = a, y = b & x = – a. The curve C encloses a plane surface lying in xy plane as shown in fig. Let us first …
Example 1 : Verify Stokes theorem for where S is the upper half surface of the sphere and C is its bounding curve. Solution : S is the surface of sphere lying above xy plane and bounded by the circle On curve So, = ydx (z = 0 on C) Consider a closed piecewise smooth …
Let S be a piecewise smooth open surface bounded by a piecewise smooth simple closed curve C. Let be a continuous vector function which has continuous first partial derivatives in a region of space which contains S in its interior. Then where C is traversed in the positive direction. The direction of C is called …
Example 1 : Let S be the surface and let be the outward unit normal to S. If then evaluate the integral . Solution : is a paraboloid with vertex at (0, 0, 1) as shown is Figure 7.36 Consider a closed surface S which consists of two piecewise smooth surface S and S’, where …
Example 1 : Evaluate where and S is the surface of paraboloid with axis parallel to z axis . Solution : The standard equation of paraboloid is given by where is the vertex of paraboloid Comparing given equation of paraboloid (Figure 7.31) with standard equation. The vertex is (0, 0, 9). Here also, we will …