# MATH 251: Calculus 3, SET8

## 16: Vector Calculus

### 16.3: The Fundamental Theorem for Line Integrals

These problems are done with the CAS. See Hand Solutions for details.

#### 1. [1094/3]

The vector field F is not conservative since it does not possess a potential function f; i.e., a function such that F, the gradient of f. In the code below, we use MATLAB's potential function to show this.
By hand, note that with F we do not have for these partial derivatives, which are continuous on the xy-plane. Thus by Theorem 5 on page 1090 of the textbook, F is not conservative.
syms x y real
P = x*y + y^2, Q = x^2 + 2*x*y
P =
Q =
F = [P Q]
F =
f = potential(F, [x y]) % NaN or Not-a-Number means DNE (does not exist) in this context.
f = NaN
dP_dy = diff(P,y), dQ_dx = diff(Q,x)
dP_dy =
dQ_dx =

#### 2. [1094/6]

The vector field F is conservative since it has a potential function f; i.e., a function such that F, the gradient of f.
(Again, see Hand Solutions for a check that this is true and how to actually construct the potential function by hand.)
syms x y
F = [y*exp(x) exp(x)+exp(y)]
F =
f = expand( potential(F, [x y]) )
f =

#### 3. [1094/8]

syms x y
F = [2*x*y + y^(-2) x^2 - 2*x*y^(-3)]
F =
f = expand( potential(F, [x y]) ) % potential function f
f =

#### 4. [1094/10]

syms x y
F = [log(y) + y/x log(x) + x/y]
F =
f = expand( potential(F, [x y]) ) % potential function f
f =

#### 5. [1094/12]

The path C is the arc of the hyperbola from to . The line integral of F along this path may be computed by the Fundamental Theorem for Line Integrals (FTLI) once we find a potential function f.
syms x y
F = [3 + 2*x*y^2 2*x^2*y]
F =
% (a)
f(x,y) = expand( potential(F, [x y]) ) % potential function f
f(x, y) =
% (b)
I = f(4,1/4) - f(1,1)
I = 9

#### 6. [1094/14]

The path C is the arc of the ellipse r, from to . The line integral of F along this path may be computed by the Fundamental Theorem for Line Integrals (FTLI) once we find a potential function f.
syms x y
F = exp(x*y) * [1+x*y x^2]
F =
% (a)
f(x,y) = expand( potential(F, [x y]) ) % potential function f
f(x, y) =
% (b)
I = f(0,2) - f(1,0)
I =

#### 7. [1095/16]

The path C is the space curve r, from to . The line integral of F along this path may be computed by the Fundamental Theorem for Line Integrals (FTLI) once we find a potential function f.
syms x y z
F = [y^2*z + 2*x*z^2 2*x*y*z x*y^2 + 2*x^2*z]
F =
% (a)
f(x,y,z) = expand( potential(F, [x y z]) ) % potential function f
f(x, y, z) =
% (b)
I = f(1,2,1) - f(0,1,0)
I = 5

#### 8. [1095/18]

The path C is the space curve r, from to . The line integral of F along this path may be computed by the Fundamental Theorem for Line Integrals (FTLI) once we find a potential function f.
syms x y z
F = [sin(y) x*cos(y) + cos(z) -y*sin(z)]
F =
% (a)
f(x,y,z) = expand( potential(F, [x y z]) ) % potential function f
f(x, y, z) =
% (b)
I = f(1,pi/2,pi) - f(0,0,0)
I =
I_appx = double(I)
I_appx = -0.5708

#### 9. [1095/20]

Here C is any path in the plane from to . The line integral of F along this path may be computed by the Fundamental Theorem for Line Integrals (FTLI) once we find a potential function f.
syms x y
F = [sin(y) x*cos(y) - sin(y)]
F =
% (a)
f(x,y) = expand( potential(F, [x y]) ) % potential function f
f(x, y) =
% (b)
I = f(1,pi) - f(2,0)
I =

#### 10. [1095/24]

Work is computed via the FTLI, analogous to #9.
syms x y
F = [2*x+y x]
F =
% (a)
f(x,y) = expand( potential(F, [x y]) ) % potential function f
f(x, y) =
% (b)
work = f(4,3) - f(1,1)
work = 26