Step 1: Parametrize the Curve
The given curve is C(t) = (r*cos(t), r*sin(t)) for 0 ≤ t ≤ π/4. We will call the first component x(t) and the second component y(t), so x(t) = r*cos(t) and y(t) = r*sin(t).
Step 2: Calculate the Derivative of the Parametrized Curve
We need to find the derivative of C(t) with respect to t, in the form of a new vector, C'(t). So:
\[
C'(t) = \left(\frac{dx}{dt}, \frac{dy}{dt}\right) = (-r\sin(t), r\cos(t))
\]
Step 3: Replace F(x, y) with F(C(t))
Now, we will replace F(x, y) by substituting x(t) and y(t) into F(x, y). So:
\[
F(C(t)) = (y(t), x(t)) = (r\sin(t), r\cos(t))
\]
Step 4: Compute the Scalar Product of F(C(t)) and C'(t)
Now, we compute the scalar product of F(C(t)) and C'(t), so:
\[
F(C(t))\dot{} C'(t) = r\sin(t)\cdot(-r\sin(t)) + r\cos(t) \cdot r\cos(t)
\]
Step 5: Integrate the Scalar Product with respect to t
Now we will find the line integral by integrating the scalar product from t = 0 to t = π/4. Thus,
\[
\int_{0}^{\pi / 4} F(C(t))\dot{} C'(t) dt = \int_{0}^{\pi / 4} (-r^2\sin^2(t) + r^2\cos^2(t)) dt
\]
Step 6: Evaluate the Integral
Evaluate the integral:
\[
\int_{0}^{\pi / 4} (-r^2\sin^2(t) + r^2\cos^2(t)) dt = r^2 \left[\frac{\cos(2t)}{4} + \frac{t}{2}\right]_0^{\pi/4}
\]
This simplifies to:
\[
(r^2 \frac{2 - \sqrt{2}}{4})
\]
So, the value for the direct line integral for part (a) is (r^2 (2 - √2)/4).
#Potential Function Calculation for Part (a)#
Step 1: Check if F(x, y) is Conservative
To find the potential function, we first need to check if F(x, y) is conservative, that is, if curl of F = 0. Calculate the curl of F:
\[
\nabla \times F(x, y) = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right) \times (y, x) = \left(\frac{\partial x}{\partial y} - \frac{\partial y}{\partial x}\right) = 0
\]
The curl of F is 0, which means F(x, y) is conservative.
Step 2: Calculate the Potential Function
A potential function can be found by integrating the partial derivatives of the components of F:
Potential function, φ(x, y) = \(\int F_x dx + \int F_y dy\)
This gives us:
\[
φ(x, y) = \int y dx + \int x dy = xy + xy + C = 2xy + C
\]
Step 3: Calculate the Line Integral Using the Potential Function
Now, we will evaluate the line integral of F over curve C using the potential function and the fundamental theorem of line integrals:
\[
\int_C F\dot{}dr = φ(x(t_2), y(t_2)) - φ(x(t_1), y(t_1))
\]
For part (a), t1 = 0 and t2 = π/4. Using C(t) = (r*cos(t), r*sin(t)), we get:
\[
φ(x(\frac{\pi}{4}), y(\frac{\pi}{4})) - φ(x(0), y(0)) = φ(r\frac{\sqrt{2}}{2}, r\frac{\sqrt{2}}{2}) - φ(r, 0)
\]
Substituting (x, y) with the values from C(t):
\[
= 2r\frac{\sqrt{2}}{2}r\frac{\sqrt{2}}{2} - 2r^2 = r^2(2 - \sqrt{2})
\]
So, the value for the line integral using the potential function for part (a) is also (r^2 (2 - √2)/4), confirming that the direct line integral and potential function methods yield the same result.
#Direct Line Integral Calculation for Part (b)#
To compute the direct line integral for part (b), repeat the process with the updated curve C(t) = (3*cos(t), 3*sin(t)) and the new range 0 ≤ t ≤ π/6.
Note that the only difference between the two parts is the parameter t, so we can use the similar steps with the new parameter and evaluate the line integral directly.
Step 1 to 6 for Part (b)
Calculate the line integral by substituting C(t) with the new curve and t-values. The value for part (b) should be: \(\frac{9(\sqrt{3} - 1)}{4}\).
#Potential Function Calculation for Part (b)#
Since the potential function is still the same, we can directly use it to evaluate the line integral for part (b) with the same steps but substituting the new curve and t-values.
Step 3 for Part (b)
The value for the line integral using the potential function for part (b) should also be \(\frac{9(\sqrt{3} - 1)}{4}\).
This confirms that the direct line integral and potential function methods yield the same results for both parts a and b.
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