Question

The domain of f(x,y) is the xy-plane, and values of f are given in the table below.

0 1 2 3 4
0 60 60 60 62 64
1 61 61 62 61 60
2 65 68 68 68 70
3 63 61 60 60 62
4 59 55 51 48 45

Find
`\int\limits_c grad f . dr` , where is
(a) A line from (0,1) to (1,2).

(b) A circle of radius 1 centered at (1,2) traversed counterclockwise.

Answer 1

a) c
`\int\limits gradf.dr` = 1

b) c
`\int\limits gradf.dr` = 0

(the limit symbol has a circle in the center for part b)

Explanation:

c
`\int\limits gradf.dr` = f(q) - f(p)

a) c
`\int\limits gradf.dr` = f(1, 2) - f(0, 1)

= 61 - 60

= 1

b) If C is the circle of radius 1 centered at the point beginning at point (1,2), we can think of C as both beginnings and ending at point ( 1, 3).

c
`\int\limits gradf.dr` = f(1, 3) - f(1, 3)

= 68 - 68

= 0

Answer 2

a. The value of the line integral along the path from (0,1) to (1,2) is 61.5.

b. The value of the circle integral around the circle of radius 1 centered at (1,2) traversed counterclockwise is 64.

Part (a): Line Integral

To calculate the line integral along the path from (0,1) to (1,2), we can parameterize the path and then use the line integral formula:

∫C f(x,y) dx dy = ∫a^b f(x(t),y(t)) |x'(t)| dt

where C is the path, x(t) and y(t) are the parametric equations of the path, and a and b are the limits of integration.

In this case, we can parameterize the path as follows:

x(t) = t

y(t) = 1 + t

where 0 ≤ t ≤ 1.

Now we can calculate the line integral:

∫C f(x,y) dx dy = ∫0^1 f(t,1+t) |1| dt

= ∫0^1 61 + t dt

= [61t + (t^2)/2]0^1 = 61.5

Therefore, the value of the line integral along the path from (0,1) to (1,2) is 61.5.

Part (b): Circle Integral

To calculate the circle integral around the circle of radius 1 centered at (1,2) traversed counterclockwise, we can use the formula:

∫C f(x,y) dx dy = ∮ f(x,y) |dx/dy| dy

where C is the circle, and x(y) and y(y) are the parametric equations of the circle.

In this case, we can parameterize the circle as follows:

x(y) = 1 + cos(2πy)

y(y) = 2 + sin(2πy)

where 0 ≤ y ≤ 1.

Now we can calculate the circle integral:

∫C f(x,y) dx dy = ∮ f(x(y),y(y)) |dx/dy| dy

= ∮ f(1+cos(2πy),2+sin(2πy)) |2πsin(2πy)| dy

= ∮ (61 + cos(2πy)) |2πsin(2πy)| dy

= ∫0^1 122πsin(2πy) + 2πcos(2πy) dy

= [61cos(2πy) + sin(2πy)]0^1 = 64

Therefore, the value of the circle integral around the circle of radius 1 centered at (1,2) traversed counterclockwise is 64.