Questions: Evaluate the line integral, where C is the given curve. ∫C xy^4 ds, C is the right half of the circle x^2+y^2=9 oriented counterclockwise

Transcript text: Evaluate the line integral, where $C$ is the given curve. $\int_{C} x y^{4} d s, C$ is the right half of the circle $x^{2}+y^{2}=9$ oriented counterclockwise

Solution

Solution Steps

Step 1: Parameterization of the Curve

The right half of the circle defined by $ x^2 + y^2 = 9 $ can be parameterized using the variable $ t $ as follows: \[ x = 3 \cos(t), \quad y = 3 \sin(t) \] for $ t $ in the interval $ [0, \pi] $.

Step 2: Calculation of the Differential Arc Length

To find the differential arc length $ ds $, we compute the derivatives of $ x $ and $ y $: \[ \frac{dx}{dt} = -3 \sin(t), \quad \frac{dy}{dt} = 3 \cos(t) \] Then, we calculate $ ds $: \[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{(-3 \sin(t))^2 + (3 \cos(t))^2} = \sqrt{9 \sin^2(t) + 9 \cos^2(t)} = 3 \]

Step 3: Expression of the Integrand

The integrand for the line integral is given by $ x y^4 $: \[ x y^4 = (3 \cos(t))(3 \sin(t))^4 = 3 \cos(t) \cdot 243 \sin^4(t) = 729 \sin^4(t) \cos(t) \]

Step 4: Setting Up the Integral

Substituting the parameterization and $ ds $ into the integral, we have: \[ \int_{C} x y^4 \, ds = \int_{0}^{\pi} 729 \sin^4(t) \cos(t) \cdot 3 \, dt = 2187 \int_{0}^{\pi} \sin^4(t) \cos(t) \, dt \]

Step 5: Evaluation of the Integral

The integral $ \int_{0}^{\pi} \sin^4(t) \cos(t) \, dt $ can be evaluated using the substitution $ u = \sin(t) $, which gives $ du = \cos(t) \, dt $. The limits change from $ t = 0 $ to $ t = \pi $ to $ u = 0 $ to $ u = 0 $. Thus, the integral evaluates to zero: \[ \int_{0}^{\pi} \sin^4(t) \cos(t) \, dt = 0 \]