Understanding the Problem: Evaluating a Line Integral

The question asks us to evaluate a line integral of a scalar function $f(x, y) = x^2y$ over a specific curve C. The curve C is defined by the parametric equations $x = 3\cos t$, $y = 3\sin t$ for the parameter range $0 \le t \le \pi/2$. This curve represents a quarter circle of radius 3 centered at the origin, starting from $(3, 0)$ and ending at $(0, 3)$. The integral is given by $\int_C f(x, y) \, dS$, where $dS$ is the differential of arc length.

Steps to Evaluate the Line Integral

To evaluate a line integral $\int_C f(x, y) \, dS$ over a curve C defined by parametric equations $x = x(t)$, $y = y(t)$ for $a \le t \le b$, we follow these steps:

1. Identify the function $f(x, y)$ and the parametric equations $x(t)$, $y(t)$ and the range of the parameter $t$.
2. Calculate the derivatives $dx/dt$ and $dy/dt$.
3. Calculate the differential of arc length $dS$ using the formula $dS = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$.
4. Substitute $x(t)$, $y(t)$, and $dS$ into the integral formula: $\int_C f(x, y) \, dS = \int_a^b f(x(t), y(t)) \, \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$.
5. Evaluate the resulting definite integral with respect to $t$.

Applying the Steps to the Given Integral

Let's apply these steps to the given problem:

Step 1: Identify the function and curve parameters

• The function is $f(x, y) = x^2y$.
• The curve C is given by $x(t) = 3\cos t$ and $y(t) = 3\sin t$.
• The parameter range is $0 \le t \le \pi/2$.

Step 2: Calculate derivatives

• $\frac{dx}{dt} = \frac{d}{dt}(3\cos t) = -3\sin t$
• $\frac{dy}{dt} = \frac{d}{dt}(3\sin t) = 3\cos t$

Step 3: Calculate dS

First, find the squares of the derivatives:

• $\left(\frac{dx}{dt}\right)^2 = (-3\sin t)^2 = 9\sin^2 t$
• $\left(\frac{dy}{dt}\right)^2 = (3\cos t)^2 = 9\cos^2 t$

Now, calculate $dS$:

$\begin{aligned} dS &= \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \\ &= \sqrt{9\sin^2 t + 9\cos^2 t} \, dt \\ &= \sqrt{9(\sin^2 t + \cos^2 t)} \, dt \end{aligned}$

Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we get:

$dS = \sqrt{9(1)} \, dt = \sqrt{9} \, dt = 3 \, dt$

Step 4: Substitute into the integral formula

Substitute $x(t)$, $y(t)$, and $dS$ into the integral $\int_C x^2y \, dS$. Remember $x = 3\cos t$ and $y = 3\sin t$:

$\begin{aligned} \int_C x^2y \, dS &= \int_0^{\pi/2} (3\cos t)^2 (3\sin t) (3 \, dt) \\ &= \int_0^{\pi/2} (9\cos^2 t) (3\sin t) (3) \, dt \\ &= \int_0^{\pi/2} 81 \cos^2 t \sin t \, dt \end{aligned}$

Step 5: Evaluate the definite integral

We need to evaluate the integral $\int_0^{\pi/2} 81 \cos^2 t \sin t \, dt$. We can use a substitution method.

Let $u = \cos t$.

Then, the differential $du = \frac{d}{dt}(\cos t) \, dt = -\sin t \, dt$. This means $\sin t \, dt = -du$.

We also need to change the limits of integration according to the substitution:

• When $t = 0$, $u = \cos(0) = 1$.
• When $t = \pi/2$, $u = \cos(\pi/2) = 0$.

Substitute $u = \cos t$ and $\sin t \, dt = -du$ and the new limits into the integral:

$\begin{aligned} \int_0^{\pi/2} 81 \cos^2 t \sin t \, dt &= \int_1^0 81 u^2 (-du) \\ &= -81 \int_1^0 u^2 \, du \end{aligned}$

Using the property $\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$, we can swap the limits and remove the negative sign:

$\begin{aligned} &= 81 \int_0^1 u^2 \, du \end{aligned}$

Now, integrate $u^2$ with respect to $u$:

$\begin{aligned} &= 81 \left[ \frac{u^{2+1}}{2+1} \right]_0^1 \\ &= 81 \left[ \frac{u^3}{3} \right]_0^1 \end{aligned}$

Evaluate the expression at the upper and lower limits:

$\begin{aligned} &= 81 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \\ &= 81 \left( \frac{1}{3} - 0 \right) \\ &= 81 \left( \frac{1}{3} \right) \\ &= \frac{81}{3} \\ &= 27 \end{aligned}$

Conclusion

The value of the integral $\int_C x^2y \, dS$, where C is the curve defined by $x = 3\cos t$, $y = 3\sin t$, $0 \le t \le \pi/2$ is 27.

Revision Table: Key Formulas for Line Integrals

Additional Information on Line Integrals

A line integral is an integral where the function to be integrated is evaluated along a curve. In this case, we evaluated a line integral of a scalar field $f(x, y)$ with respect to arc length ($dS$). The term $dS$ represents an infinitesimally small piece of the curve's length. When evaluating the integral, we transform it into a definite integral with respect to the parameter $t$, using the parametrization of the curve.

The substitution method used in the final step is a standard technique for evaluating integrals. By letting $u$ be a function of $t$, we can transform the integral with respect to $t$ into an integral with respect to $u$, often simplifying the integration process. It's crucial to remember to change the limits of integration according to the substitution when evaluating definite integrals.