Part B: 10-Mark Questions
1. (i) Find the angle between two vectors x2+y2+z2-9=0 and x2+y2-z-3=0 at (2,-1,2).
Sol: The angle between surfaces is the angle between their normal vectors, given by the gradient (∇φ).
Let φ1 = x2 + y2 + z2 - 9
∇φ1 = (∂/∂x)\(\vec{i}\) + (∂/∂y)\(\vec{j}\) + (∂/∂z)\(\vec{k}\)
∇φ1 = 2x \(\vec{i}\) + 2y \(\vec{j}\) + 2z \(\vec{k}\)
At (2, -1, 2), \(\vec{N}_1\) = ∇φ1 = 2(2)\(\vec{i}\) + 2(-1)\(\vec{j}\) + 2(2)\(\vec{k}\) = 4\(\vec{i}\) - 2\(\vec{j}\) + 4\(\vec{k}\)
Let φ2 = x2 + y2 - z - 3
∇φ2 = 2x \(\vec{i}\) + 2y \(\vec{j}\) - 1 \(\vec{k}\)
At (2, -1, 2), \(\vec{N}_2\) = ∇φ2 = 2(2)\(\vec{i}\) + 2(-1)\(\vec{j}\) - 1\(\vec{k}\) = 4\(\vec{i}\) - 2\(\vec{j}\) - \(\vec{k}\)
We use the dot product: \(\vec{N}_1\) · \(\vec{N}_2\) = |\(\vec{N}_1\)| |\(\vec{N}_2\)| cos θ
\(\vec{N}_1\) · \(\vec{N}_2\) = (4)(4) + (-2)(-2) + (4)(-1) = 16 + 4 - 4 = 16
|\(\vec{N}_1\)| = √(42 + (-2)2 + 42) = √(16 + 4 + 16) = √36 = 6
|\(\vec{N}_2\)| = √(42 + (-2)2 + (-1)2) = √(16 + 4 + 1) = √21
cos θ = (\(\vec{N}_1\) · \(\vec{N}_2\)) / (|\(\vec{N}_1\)| |\(\vec{N}_2\)|) = 16 / (6 √21) = 8 / (3√21)
θ = cos-1(8 / (3√21))
(ii) If \(\vec{F}\) = xz3 \(\vec{i}\) - 2xyz \(\vec{j}\) + xz \(\vec{k}\). Find div \(\vec{F}\) and curl \(\vec{F}\) at (1,2,0).
Sol:
Divergence (div \(\vec{F}\)):
div \(\vec{F}\) = ∇ · \(\vec{F}\) = (∂/∂x)(xz3) + (∂/∂y)(-2xyz) + (∂/∂z)(xz)
= z3 - 2xz + x
At (1, 2, 0): div \(\vec{F}\) = (0)3 - 2(1)(0) + 1 = 1
Curl (curl \(\vec{F}\)):
curl \(\vec{F}\) = ∇ × \(\vec{F}\) =
| \(\vec{i}\) \(\vec{j}\) \(\vec{k}\) |
| ∂/∂x ∂/∂y ∂/∂z |
| xz3 -2xyz xz |
= \(\vec{i}\) [(∂/∂y)(xz) - (∂/∂z)(-2xyz)] - \(\vec{j}\) [(∂/∂x)(xz) - (∂/∂z)(xz3)] + \(\vec{k}\) [(∂/∂x)(-2xyz) - (∂/∂y)(xz3)]
= \(\vec{i}\) [0 - (-2xy)] - \(\vec{j}\) [z - 3xz2] + \(\vec{k}\) [-2yz - 0]
= (2xy) \(\vec{i}\) - (z - 3xz2) \(\vec{j}\) - (2yz) \(\vec{k}\)
At (1, 2, 0): curl \(\vec{F}\) = (2(1)(2)) \(\vec{i}\) - (0 - 3(1)(0)2) \(\vec{j}\) - (2(2)(0)) \(\vec{k}\)
= 4\(\vec{i}\) - 0\(\vec{j}\) - 0\(\vec{k}\) = 4\(\vec{i}\)
2. (i) Show that surfaces φ=y2+z2-x and ψ=log(y2+z2)+4x are orthogonal.
Sol: Two surfaces are orthogonal if the dot product of their normal vectors (gradients) is zero.
\(\vec{N}_1\) = ∇φ = (∂φ/∂x)\(\vec{i}\) + (∂φ/∂y)\(\vec{j}\) + (∂φ/∂z)\(\vec{k}\)
\(\vec{N}_1\) = -1\(\vec{i}\) + 2y\(\vec{j}\) + 2z\(\vec{k}\)
\(\vec{N}_2\) = ∇ψ = (∂ψ/∂x)\(\vec{i}\) + (∂ψ/∂y)\(\vec{j}\) + (∂ψ/∂z)\(\vec{k}\)
\(\vec{N}_2\) = 4\(\vec{i}\) + (2y / (y2+z2))\(\vec{j}\) + (2z / (y2+z2))\(\vec{k}\)
Now, find the dot product \(\vec{N}_1\) · \(\vec{N}_2\):
\(\vec{N}_1\) · \(\vec{N}_2\) = (-1)(4) + (2y)(2y / (y2+z2)) + (2z)(2z / (y2+z2))
= -4 + 4y2 / (y2+z2) + 4z2 / (y2+z2)
= -4 + (4y2 + 4z2) / (y2+z2)
= -4 + 4(y2 + z2) / (y2+z2)
= -4 + 4 = 0
Since ∇φ · ∇ψ = 0, the surfaces are orthogonal.
(ii) Find constants a,b,c so that \(\vec{F}\) = (x+2y+az)\(\vec{i}\) + (bx-3y-z)\(\vec{j}\) + (4x+cy+2z)\(\vec{k}\) is irrotational.
Sol: A vector \(\vec{F}\) is irrotational if its curl is zero (curl \(\vec{F}\) = \(\vec{0}\)).
curl \(\vec{F}\) = ∇ × \(\vec{F}\) =
| \(\vec{i}\) \(\vec{j}\) \(\vec{k}\) |
| ∂/∂x ∂/∂y ∂/∂z |
| x+2y+az bx-3y-z 4x+cy+2z |
= \(\vec{i}\) [(∂/∂y)(4x+cy+2z) - (∂/∂z)(bx-3y-z)] - \(\vec{j}\) [(∂/∂x)(4x+cy+2z) - (∂/∂z)(x+2y+az)] + \(\vec{k}\) [(∂/∂x)(bx-3y-z) - (∂/∂y)(x+2y+az)]
= \(\vec{i}\) [c - (-1)] - \(\vec{j}\) [4 - a] + \(\vec{k}\) [b - 2]
= (c+1)\(\vec{i}\) - (4-a)\(\vec{j}\) + (b-2)\(\vec{k}\)
For \(\vec{F}\) to be irrotational, curl \(\vec{F}\) = 0\(\vec{i}\) + 0\(\vec{j}\) + 0\(\vec{k}\).
Equating components:
c + 1 = 0 ⇒ c = -1
4 - a = 0 ⇒ a = 4
b - 2 = 0 ⇒ b = 2
The constants are a=4, b=2, c=-1.
3. Verify Gauss divergence theorem for \(\vec{F}\) = x\(\vec{i}\) + y\(\vec{j}\) + z\(\vec{k}\) over the cube bounded by x=0, x=a, y=0, y=a, z=0, z=a.
Sol: Gauss Divergence Theorem states: ∫∫S (\(\vec{F}\) · \(\vec{n}\)) dS = ∫∫∫V (∇ · \(\vec{F}\)) dV
RHS (Volume Integral):
∇ · \(\vec{F}\) = (∂/∂x)(x) + (∂/∂y)(y) + (∂/∂z)(z) = 1 + 1 + 1 = 3
∫∫∫V (∇ · \(\vec{F}\)) dV = ∫0a ∫0a ∫0a 3 dx dy dz
= 3 × (Volume of cube) = 3a3
LHS (Surface Integral): The cube has 6 faces.
1. Face 1 (x=a): \(\vec{n}\) = \(\vec{i}\), dS = dy dz. \(\vec{F}\) = a\(\vec{i}\) + y\(\vec{j}\) + z\(\vec{k}\).
∫∫ (\(\vec{F}\) · \(\vec{n}\)) dS = ∫0a ∫0a (a) dy dz = a(a)(a) = a3
2. Face 2 (x=0): \(\vec{n}\) = -\(\vec{i}\), dS = dy dz. \(\vec{F}\) = 0\(\vec{i}\) + y\(\vec{j}\) + z\(\vec{k}\).
∫∫ (\(\vec{F}\) · \(\vec{n}\)) dS = ∫0a ∫0a (0) dy dz = 0
3. Face 3 (y=a): \(\vec{n}\) = \(\vec{j}\), dS = dx dz. \(\vec{F}\) = x\(\vec{i}\) + a\(\vec{j}\) + z\(\vec{k}\).
∫∫ (\(\vec{F}\) · \(\vec{n}\)) dS = ∫0a ∫0a (a) dx dz = a(a)(a) = a3
4. Face 4 (y=0): \(\vec{n}\) = -\(\vec{j}\), dS = dx dz. \(\vec{F}\) = x\(\vec{i}\) + 0\(\vec{j}\) + z\(\vec{k}\).
∫∫ (\(\vec{F}\) · \(\vec{n}\)) dS = ∫0a ∫0a (0) dx dz = 0
5. Face 5 (z=a): \(\vec{n}\) = \(\vec{k}\), dS = dx dy. \(\vec{F}\) = x\(\vec{i}\) + y\(\vec{j}\) + a\(\vec{k}\).
∫∫ (\(\vec{F}\) · \(\vec{n}\)) dS = ∫0a ∫0a (a) dx dy = a(a)(a) = a3
6. Face 6 (z=0): \(\vec{n}\) = -\(\vec{k}\), dS = dx dy. \(\vec{F}\) = x\(\vec{i}\) + y\(\vec{j}\) + 0\(\vec{k}\).
∫∫ (\(\vec{F}\) · \(\vec{n}\)) dS = ∫0a ∫0a (0) dx dy = 0
Total Surface Integral (LHS) = a3 + 0 + a3 + 0 + a3 + 0 = 3a3
Since LHS (3a3) = RHS (3a3), the theorem is verified.
4. Evaluate ∫∫S \(\vec{F}\) · \(\vec{n}\) dS where \(\vec{F}\) = (y-z+2)\(\vec{i}\) + (y+4)\(\vec{j}\) - xz\(\vec{k}\) where S is the cube bounded by x=0, x=2, y=0, y=2, z=0, z=2.
Sol: By Gauss Divergence Theorem, ∫∫S (\(\vec{F}\) · \(\vec{n}\)) dS = ∫∫∫V (∇ · \(\vec{F}\)) dV
First, find the divergence of \(\vec{F}\):
∇ · \(\vec{F}\) = (∂/∂x)(y-z+2) + (∂/∂y)(y+4) + (∂/∂z)(-xz)
= 0 + 1 + (-x)
= 1 - x
Now, compute the volume integral over the cube:
V = ∫∫∫V (1 - x) dV = ∫02 ∫02 ∫02 (1 - x) dx dy dz
We can separate the integrals:
V = [ ∫02 dz ] × [ ∫02 dy ] × [ ∫02 (1 - x) dx ]
∫02 dz = [z]02 = 2
∫02 dy = [y]02 = 2
∫02 (1 - x) dx = [x - x2/2]02 = (2 - 22/2) - (0) = (2 - 4/2) = 2 - 2 = 0
V = (2) × (2) × (0) = 0
The value of the surface integral is 0.
5. Verify Gauss divergence theorem for \(\vec{F}\) = x2\(\vec{i}\) + y2\(\vec{j}\) + z2\(\vec{k}\) over the cuboid bounded by x=0, x=b, y=0, y=b, z=0, z=c.
Sol: Gauss Divergence Theorem states: ∫∫S (\(\vec{F}\) · \(\vec{n}\)) dS = ∫∫∫V (∇ · \(\vec{F}\)) dV
RHS (Volume Integral):
∇ · \(\vec{F}\) = (∂/∂x)(x2) + (∂/∂y)(y2) + (∂/∂z)(z2) = 2x + 2y + 2z
∫∫∫V (2x + 2y + 2z) dV = ∫0c ∫0b ∫0b (2x + 2y + 2z) dx dy dz
Inner (dx): ∫0b (2x + 2y + 2z) dx = [x2 + 2yx + 2zx]0b = b2 + 2yb + 2zb
Middle (dy): ∫0b (b2 + 2yb + 2zb) dy = [b2y + y2b + 2zby]0b = (b3 + b3 + 2zb2) = 2b3 + 2zb2
Outer (dz): ∫0c (2b3 + 2zb2) dz = [2b3z + z2b2]0c = 2b3c + c2b2
RHS = b2c(2b + c)
LHS (Surface Integral):
1. Face 1 (x=b): \(\vec{n}\) = \(\vec{i}\), dS = dy dz. \(\vec{F}\) = b2\(\vec{i}\) + y2\(\vec{j}\) + z2\(\vec{k}\). ∫∫ (b2) dy dz = ∫0c ∫0b b2 dy dz = b2(b)(c) = b3c
2. Face 2 (x=0): \(\vec{n}\) = -\(\vec{i}\). \(\vec{F}\) = 0\(\vec{i}\) + y2\(\vec{j}\) + z2\(\vec{k}\). ∫∫ (0) dy dz = 0
3. Face 3 (y=b): \(\vec{n}\) = \(\vec{j}\), dS = dx dz. \(\vec{F}\) = x2\(\vec{i}\) + b2\(\vec{j}\) + z2\(\vec{k}\). ∫∫ (b2) dx dz = ∫0c ∫0b b2 dx dz = b2(b)(c) = b3c
4. Face 4 (y=0): \(\vec{n}\) = -\(\vec{j}\). \(\vec{F}\) = x2\(\vec{i}\) + 0\(\vec{j}\) + z2\(\vec{k}\). ∫∫ (0) dx dz = 0
5. Face 5 (z=c): \(\vec{n}\) = \(\vec{k}\), dS = dx dy. \(\vec{F}\) = x2\(\vec{i}\) + y2\(\vec{j}\) + c2\(\vec{k}\). ∫∫ (c2) dx dy = ∫0b ∫0b c2 dx dy = c2(b)(b) = c2b2
6. Face 6 (z=0): \(\vec{n}\) = -\(\vec{k}\). \(\vec{F}\) = x2\(\vec{i}\) + y2\(\vec{j}\) + 0\(\vec{k}\). ∫∫ (0) dx dy = 0
Total Surface Integral (LHS) = b3c + 0 + b3c + 0 + c2b2 + 0 = 2b3c + c2b2 = b2c(2b + c)
Since LHS = RHS, the theorem is verified.
6. Verify Stoke’s theorem for \(\vec{F}\) = (x2-y2)\(\vec{i}\) + 2xy\(\vec{j}\) in the rectangular region x=0, x=a, y=0, y=b.
Sol: Stoke's Theorem states: ∮C (\(\vec{F}\) · d\(\vec{r}\)) = ∫∫S (curl \(\vec{F}\)) · \(\vec{n}\) dS
Region is in the z=0 plane, so \(\vec{n}\) = \(\vec{k}\) and dS = dx dy.
RHS (Surface Integral):
curl \(\vec{F}\) = ∇ × \(\vec{F}\) =
| \(\vec{i}\) \(\vec{j}\) \(\vec{k}\) |
| ∂/∂x ∂/∂y ∂/∂z |
| x2-y2 2xy 0 |
= \(\vec{i}\)(0-0) - \(\vec{j}\)(0-0) + \(\vec{k}\)[(∂/∂x)(2xy) - (∂/∂y)(x2-y2)]
= \(\vec{k}\)[2y - (-2y)] = 4y\(\vec{k}\)
(curl \(\vec{F}\)) · \(\vec{n}\) = (4y\(\vec{k}\)) · (\(\vec{k}\)) = 4y
∫∫S (4y) dS = ∫0a ∫0b 4y dy dx
= ( ∫0a dx ) × ( ∫0b 4y dy )
= [x]0a × [2y2]0b = (a) × (2b2) = 2ab2
LHS (Line Integral): ∮C \(\vec{F}\) · d\(\vec{r}\) = ∮C (x2-y2)dx + 2xy dy
Path is O(0,0) → A(a,0) → B(a,b) → C(0,b) → O(0,0)
1. C1 (O→A): y=0, dy=0. x from 0 to a. ∫0a (x2-0)dx = [x3/3]0a = a3/3
2. C2 (A→B): x=a, dx=0. y from 0 to b. ∫0b 2(a)y dy = [ay2]0b = ab2
3. C3 (B→C): y=b, dy=0. x from a to 0. ∫a0 (x2-b2)dx = [x3/3 - b2x]a0 = (0) - (a3/3 - ab2) = -a3/3 + ab2
4. C4 (C→O): x=0, dx=0. y from b to 0. ∫b0 0 dy = 0
Total Line Integral (LHS) = (a3/3) + (ab2) + (-a3/3 + ab2) + 0 = 2ab2
Since LHS = RHS, the theorem is verified.
7. Verify Stoke’s theorem for \(\vec{F}\) = x2\(\vec{i}\) + xy\(\vec{j}\) integrated round the square z=0, x=0, y=0, x=a, y=a.
Sol: Stoke's Theorem states: ∮C (\(\vec{F}\) · d\(\vec{r}\)) = ∫∫S (curl \(\vec{F}\)) · \(\vec{n}\) dS
Region is in z=0 plane, so \(\vec{n}\) = \(\vec{k}\) and dS = dx dy.
RHS (Surface Integral):
curl \(\vec{F}\) = ∇ × \(\vec{F}\) =
| \(\vec{i}\) \(\vec{j}\) \(\vec{k}\) |
| ∂/∂x ∂/∂y ∂/∂z |
| x2 xy 0 |
= \(\vec{i}\)(0-0) - \(\vec{j}\)(0-0) + \(\vec{k}\)[(∂/∂x)(xy) - (∂/∂y)(x2)]
= \(\vec{k}\)[y - 0] = y\(\vec{k}\)
(curl \(\vec{F}\)) · \(\vec{n}\) = (y\(\vec{k}\)) · (\(\vec{k}\)) = y
∫∫S (y) dS = ∫0a ∫0a y dy dx
= ( ∫0a dx ) × ( ∫0a y dy )
= [x]0a × [y2/2]0a = (a) × (a2/2) = a3/2
LHS (Line Integral): ∮C \(\vec{F}\) · d\(\vec{r}\) = ∮C x2dx + xy dy
Path is O(0,0) → A(a,0) → B(a,a) → C(0,a) → O(0,0)
1. C1 (O→A): y=0, dy=0. x from 0 to a. ∫0a x2dx = [x3/3]0a = a3/3
2. C2 (A→B): x=a, dx=0. y from 0 to a. ∫0a (a)y dy = [ay2/2]0a = a3/2
3. C3 (B→C): y=a, dy=0. x from a to 0. ∫a0 x2dx = [x3/3]a0 = 0 - (a3/3) = -a3/3
4. C4 (C→O): x=0, dx=0. y from a to 0. ∫a0 0 dy = 0
Total Line Integral (LHS) = (a3/3) + (a3/2) + (-a3/3) + 0 = a3/2
Since LHS = RHS, the theorem is verified.
8. Verify Green’s theorem for ∮C [(x2-y2)dx + 4xydy] in the rectangular region x=0, x=a, y=0, y=b.
Sol: Green's Theorem states: ∮C (P dx + Q dy) = ∫∫R ( (∂Q/∂x) - (∂P/∂y) ) dA
Here, P = x2-y2, Q = 4xy, dA = dx dy
RHS (Double Integral):
∂Q/∂x = 4y
∂P/∂y = -2y
(∂Q/∂x) - (∂P/∂y) = 4y - (-2y) = 6y
∫∫R (6y) dA = ∫0a ∫0b 6y dy dx
= ( ∫0a dx ) × ( ∫0b 6y dy )
= [x]0a × [3y2]0b = (a) × (3b2) = 3ab2
LHS (Line Integral): Path O(0,0) → A(a,0) → B(a,b) → C(0,b) → O(0,0)
1. C1 (O→A): y=0, dy=0. x from 0 to a. ∫0a (x2-0)dx = [x3/3]0a = a3/3
2. C2 (A→B): x=a, dx=0. y from 0 to b. ∫0b 4(a)y dy = [2ay2]0b = 2ab2
3. C3 (B→C): y=b, dy=0. x from a to 0. ∫a0 (x2-b2)dx = [x3/3 - b2x]a0 = (0) - (a3/3 - ab2) = -a3/3 + ab2
4. C4 (C→O): x=0, dx=0. y from b to 0. ∫b0 0 dy = 0
Total Line Integral (LHS) = (a3/3) + (2ab2) + (-a3/3 + ab2) + 0 = 3ab2
Since LHS = RHS, the theorem is verified.
9. Verify Green’s theorem for ∮C [(xy+y2)dx + x2dy] where C is bounded by y=x and y=x2.
Sol: Green's Theorem: ∮C (P dx + Q dy) = ∫∫R ( (∂Q/∂x) - (∂P/∂y) ) dA
Here, P = xy+y2, Q = x2. Intersection points are (0,0) and (1,1).
RHS (Double Integral):
∂Q/∂x = 2x
∂P/∂y = x + 2y
(∂Q/∂x) - (∂P/∂y) = 2x - (x + 2y) = x - 2y
∫∫R (x - 2y) dA = ∫01 ∫x2x (x - 2y) dy dx
Inner (dy): ∫x2x (x - 2y) dy = [xy - y2]y=x2y=x
= (x(x) - x2) - (x(x2) - (x2)2)
= (x2 - x2) - (x3 - x4) = -x3 + x4
Outer (dx): ∫01 (-x3 + x4) dx = [-x4/4 + x5/5]01
= (-1/4 + 1/5) - 0 = -5/20 + 4/20 = -1/20
LHS (Line Integral):
1. C1 (y=x2, from (0,0) to (1,1)): dy = 2x dx. x from 0 to 1.
∫ (xy+y2)dx + x2dy = ∫01 (x(x2)+(x2)2)dx + (x2)(2x dx)
= ∫01 (x3 + x4 + 2x3) dx = ∫01 (3x3 + x4) dx
= [3x4/4 + x5/5]01 = (3/4 + 1/5) = 19/20
2. C2 (y=x, from (1,1) to (0,0)): dy = dx. x from 1 to 0.
∫ (xy+y2)dx + x2dy = ∫10 (x(x) + x2)dx + x2(dx)
= ∫10 (x2 + x2 + x2) dx = ∫10 3x2 dx
= [x3]10 = 0 - 1 = -1
Total Line Integral (LHS) = 19/20 + (-1) = 19/20 - 20/20 = -1/20
Since LHS = RHS, the theorem is verified.
10. Verify Green’s theorem for ∮C [(3x2-8y2)dx + (4y-6xy)dy]. C is bounded by y=√x and y=x2.
Sol: Green's Theorem: ∮C (P dx + Q dy) = ∫∫R ( (∂Q/∂x) - (∂P/∂y) ) dA
Here, P = 3x2-8y2, Q = 4y-6xy. Intersection points are (0,0) and (1,1).
RHS (Double Integral):
∂Q/∂x = -6y
∂P/∂y = -16y
(∂Q/∂x) - (∂P/∂y) = -6y - (-16y) = 10y
∫∫R (10y) dA = ∫01 ∫x2√x 10y dy dx
Inner (dy): ∫x2√x 10y dy = [5y2]y=x2y=√x
= 5[ (√x)2 - (x2)2 ] = 5(x - x4)
Outer (dx): ∫01 5(x - x4) dx = 5 [x2/2 - x5/5]01
= 5 [ (1/2 - 1/5) - 0 ] = 5 [ 5/10 - 2/10 ] = 5(3/10) = 3/2
LHS (Line Integral):
1. C1 (y=x2, from (0,0) to (1,1)): dy = 2x dx. x from 0 to 1.
∫ (3x2-8y2)dx + (4y-6xy)dy
= ∫01 (3x2-8(x2)2)dx + (4(x2)-6x(x2))(2x dx)
= ∫01 (3x2-8x4)dx + (4x2-6x3)(2x dx)
= ∫01 (3x2-8x4 + 8x3-12x4) dx = ∫01 (3x2 + 8x3 - 20x4) dx
= [x3 + 2x4 - 4x5]01 = (1 + 2 - 4) - 0 = -1
2. C2 (y=√x or x=y2, from (1,1) to (0,0)): dx = 2y dy. y from 1 to 0.
∫ (3x2-8y2)dx + (4y-6xy)dy
= ∫10 (3(y2)2-8y2)(2y dy) + (4y-6(y2)y)dy
= ∫10 (3y4-8y2)(2y)dy + (4y-6y3)dy
= ∫10 (6y5 - 16y3 + 4y - 6y3) dy = ∫10 (6y5 - 22y3 + 4y) dy
= [y6 - 22y4/4 + 2y2]10 = [y6 - (11/2)y4 + 2y2]10
= (0) - (1 - 11/2 + 2) = -(3 - 11/2) = -(6/2 - 11/2) = -(-5/2) = 5/2
Total Line Integral (LHS) = -1 + 5/2 = -2/2 + 5/2 = 3/2
Since LHS = RHS, the theorem is verified.