The first such Integral is a Line Integral which is an integral specified over a curve. It is possible to have line integrals for both scalar and vector fields. For this course we will mostly focus on the Line Integral for vector fields. An example is WORK in Physics which is ∫ C F.dr where F is the Force Vector( field) and dr is the displacement vector ( field) and the path is over the curve C from points a to point b. The line integral satisfies the usual definition of an integral as the limit of a sum. In general the value of the line integral is dependent on the path C except for a special class of vector fields which may be expressed as gradient of a scalar field. In this special case the line integral becomes path independent and dependent only on the values at the two boundary points.
The line integral over a closed curve is called the CIRCULATION of the vector field.
Sum of circulations over a number of small closed curves making up a larger close curve is equal to the circulation over the larger curve.
The second such integral is the surface integral which is an integral over a specified surface S which may be OPEN or CLOSED surface. An open surface has a
boundary like the surface of a book. The boundary of an OPEN surface is a curve.
( Many open surfaces may have a common boundary). Direction of positive normal vector to an open surface is ambiguous and we have to make a choice for the positive direction. A Closed surface has no boundary and has an INSIDE and an OUTSIDE. The outer normal vector is taken to be positive for closed surfaces. ( Remember that
area is a vector, actually a pseudovector and its direction is normal to the area element). Outward normal is taken to be positive.
A surface integral may also be understood as the limit of a sum. Surface integrals are dependent on the choice of the surface S except for a special class of vector fields. Surface integrals of vector fields gives the FLUX of the vector field over the surface specified. Just like in Circulations the FLUX of vector field over a
closed surface is given by the flux through subdivision of the surface to smaller surfaces as adjacent surface contributions cancel out.
Last is the Volume Integral which is an integral of either a scalar field or a vector field over a specified volume V. ( Remember that volume is a scalar, actually a pseudo scalar).
Finally we covered the first boundary theorem for the gradient of a scalar field.
The line integral of the gradient of a scalar field was found to be path independent
and dependent only on the end points ( boundary points) of the curve. Remember that two points can be connected by an infinite number of OPEN curves. Over all these the line integral of the gradient of a scalar field will have the same value given by the difference in the value of the scalar field at the two end points
φ (b)- φ (a)
sir,in the class while talking about the flux through an infinitesimal cube,you took flux through the surface 1, Cx(1),as constant because we move an infinitesimal distance along y-z plane, and took it out of the integral, but while calculating the flux through the 2nd surface,why is taylor's theorem used,even if we move a very small distance ,dx(while moving from surface1 to surface2).Cx(1) should have been equal to Cx(2) because they are separated by an infinitesimal distance.
ReplyDeleteThis issue was raised in a question during the class
ReplyDeleteand i explained it. You need to be more attentive in class.
The reason is as follows; we consider the vector field C(x, y, z) constant over dydz but varying
along dx because if you do a careful Taylor expansion you will see that C changes only at second order infintesimal i.e. proportional to
(dy) square or (dz) square which are of order zero. ( Remember we are working only to first order in infinitesimal quantities). Along (dx) however C changes right from first order. This is the reason
we take C to be constant over (dy dz) because the change is second order (of order zero) but consider the chnage in C over dx which is of first order ( not of order zero).
We use the same logic also in proving stokes theorem.
sir, are all vectors of the type
ReplyDelete^i f(x,y,z)+^j g(x,y,z)+^k h(x,y,z)
conservative vectors with respect to line integration?
if no, then what type of vectors are conservative ?
please elaborate...
sir i didn't understand stoke's theorem , please explain sir
ReplyDeletesir, are all vectors of with scalar components of the type f(x,y,z) conservative with respect to line integral ?
ReplyDeleteif not then please elaborate on the type of vectors which are conservative .
No. That is just a general vector field each of whose component is a scalar field. Only vector fields whose line integrals are path independent or
ReplyDeleteequivalently from the First boundary theorem on gradients, any vector field v(x,y,z) such that
v is expressible as the gradient of some scalar field phi (x, y, z) or v=grad phi
If you dont understand in class please stop me there
ReplyDeleteso that i cam explain. It is important in the semester system to understand in the class itself. For this you need to do regular studies of each subject.
I will be putting an update on Stokes theorem and Gauss theorem soon on this blog. Check it when it is updated and then if you still dont understand you should contact me after the lectures on Monday.
One question on conservative vector fields is repeated. I have already answered it, please check the comments on this post carefully.
ReplyDelete