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Answer :
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.
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Rewritten by : Barada
Answer:
a) c[tex]\int\limits gradf.dr[/tex] = 1
b) c[tex]\int\limits gradf.dr[/tex] = 0
(the limit symbol has a circle in the center for part b)
Step-by-step explanation:
c[tex]\int\limits gradf.dr[/tex] = f(q) - f(p)
a) c[tex]\int\limits gradf.dr[/tex] = 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[tex]\int\limits gradf.dr[/tex] = f(1, 3) - f(1, 3)
= 68 - 68
= 0
