-
If A and B represent the endpoints of the path C, and f is a scalar function such that F is the gradient of f, then the fundamental theorem of line integrals states that
∫ F · dr = f(B) - f(A),
which is nearly like the fundamental theorem of calculus! (Think of f as an "antigradient" of F.)
This theorem tells us that the line integral depends only on the endpoints and not on the path taken, if F possesses an "antigradient" f (i.e. if F is a conservative vector field).
The first step will be to find such an f (if such an f exists). The fact that we are told to use this fundamental theorem likely means that in this particular problem, such an f exists.
grad f = F leads to the following three equations:
1) df/dx = 2x+yz
2) df/dy = 2y+xz
3) df/dz = 4z+xy.
Integrating 1) gives f(x,y,z) = x^2+xyz+g(y,z) where g is an arbitrary function of only y and z.
Substituting into 2) gives xz+dg/dy = 2y+xz, so g(y,z) = y^2+h(z), where h is an arbitrary function of only z. So f(x,y,z) = x^2+xyz+y^2+h(z).
Substituting into 3) gives xy+h'(z) = 4z+xy, so h(z) = 2z^2+c.
Thus, f(x,y,z) = x^2+xyz+y^2+2z^2+c.
The rest is easy. We find the endpoints A and B by plugging in the initial and final values of t, and then we apply the fundamental theorem by evaluating f(B) - f(A). Note that the +c terms will cancel.
A = (x(0), y(0), z(0)) = (3(0)+1, 4+2cos(0π/2), 0^2 - 1) = (1,6,-1).
B = (x(2), y(2), z(2)) = (3(2)+1, 4+2cos(2π/2), 2^2 - 1) = (7,2,3).
(So a mistake was made in the statement of the problem. The path is from (1,6,-1) to (7,2,3), not to (4,2,3), since 3t+1 is 7 and not 4 when t=2.)
So ∫ F · dr = f(B) - f(A)
= f(7,2,3) - f(1,6,-1)
= 49+42+4+18+c - (1-6+36+2+c)
= 113+c - (33+c)
= 113+c-33-c
= 80.
Lord bless you today!
∫ F · dr = f(B) - f(A),
which is nearly like the fundamental theorem of calculus! (Think of f as an "antigradient" of F.)
This theorem tells us that the line integral depends only on the endpoints and not on the path taken, if F possesses an "antigradient" f (i.e. if F is a conservative vector field).
The first step will be to find such an f (if such an f exists). The fact that we are told to use this fundamental theorem likely means that in this particular problem, such an f exists.
grad f = F leads to the following three equations:
1) df/dx = 2x+yz
2) df/dy = 2y+xz
3) df/dz = 4z+xy.
Integrating 1) gives f(x,y,z) = x^2+xyz+g(y,z) where g is an arbitrary function of only y and z.
Substituting into 2) gives xz+dg/dy = 2y+xz, so g(y,z) = y^2+h(z), where h is an arbitrary function of only z. So f(x,y,z) = x^2+xyz+y^2+h(z).
Substituting into 3) gives xy+h'(z) = 4z+xy, so h(z) = 2z^2+c.
Thus, f(x,y,z) = x^2+xyz+y^2+2z^2+c.
The rest is easy. We find the endpoints A and B by plugging in the initial and final values of t, and then we apply the fundamental theorem by evaluating f(B) - f(A). Note that the +c terms will cancel.
A = (x(0), y(0), z(0)) = (3(0)+1, 4+2cos(0π/2), 0^2 - 1) = (1,6,-1).
B = (x(2), y(2), z(2)) = (3(2)+1, 4+2cos(2π/2), 2^2 - 1) = (7,2,3).
(So a mistake was made in the statement of the problem. The path is from (1,6,-1) to (7,2,3), not to (4,2,3), since 3t+1 is 7 and not 4 when t=2.)
So ∫ F · dr = f(B) - f(A)
= f(7,2,3) - f(1,6,-1)
= 49+42+4+18+c - (1-6+36+2+c)
= 113+c - (33+c)
= 113+c-33-c
= 80.
Lord bless you today!
-
It's fairly obvious from inspection that
F = ∇f
where f = x² + y² + 2z² + xyz
So now the FToLI says that
∫C F • dr = f(r(2)) - f(r(0)) = f(4,2,3) - f(1,6,-1)
although I suspect there's a typo somewhere because r(2) ≠ (4,2,3) since x(2) = 7, but the fact there's a potential function is the key.
F = ∇f
where f = x² + y² + 2z² + xyz
So now the FToLI says that
∫C F • dr = f(r(2)) - f(r(0)) = f(4,2,3) - f(1,6,-1)
although I suspect there's a typo somewhere because r(2) ≠ (4,2,3) since x(2) = 7, but the fact there's a potential function is the key.