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line integrals with ds
integral of f(r(t)) * ||r’(t)||dt
where r(t) is the parameterization of the given line, and can be found using (1-t)<a,b,c> + t<e,f,g>

line integrals of vector fields
given vector field F(p,q,w) and your vector r(t), fill out this integral
integral from bounds of the line in terms of t of p(r(t))*x’(t) + q(r(t))*y’(t) + w(r(t))*z’(t) dt
basically take the variables in the p slot, for example p = 1+6xy and that becomes 1 +6x(t)y(t), you turn all those variables into functions of t, and then you use the parameterization to sub out it all in terms of t, do a big fat reduction and solve for the integral
think of it like the particle is going in a path, vector field is force and the line integral of vector field is the work done on it. positive integral mean positive work done, negative integral means force fights against object.
ds with a double integral
|| ru x rv || dA
conversion jacobians
always do the jacobian when doing u and v subs, but when switching direct its just
polar → rd(theta)d(phi)
spherical → roh2sin(phi)d(theta)d(phi)d(roh)