Questions: Find the area enclosed by the curve (x=6 sin t), (y=e^-1/3), (0 leq t leq pi) and the (y)-axis. Write the exact answer. Do not round.

Transcript text: Find the area enclosed by the curve $x=6 \operatorname{sint}, y=e^{-\frac{1}{3}}, 0 \leq t \leq \pi$ and the $y$-axis. Write the exact answer. Do not round.

Solution

Solution Steps

To find the area enclosed by the given parametric curve and the y-axis, we need to set up an integral. The curve is defined by $ x = 6 \sin(t) $ and $ y = e^{-\frac{1}{3}} $ for $ 0 \leq t \leq \pi $. The area can be found by integrating the x-component with respect to $ t $ over the given interval.

Solution Approach

Identify the parametric equations: $ x = 6 \sin(t) $ and $ y = e^{-\frac{1}{3}} $.
Set up the integral for the area using the formula for the area under a parametric curve: $ \int_{a}^{b} y \frac{dx}{dt} dt $.
Compute $ \frac{dx}{dt} $.
Integrate $ y \frac{dx}{dt} $ over the interval $ 0 \leq t \leq \pi $.

Step 1: Define the Parametric Equations

The parametric equations for the curve are given by: \[ x = 6 \sin(t), \quad y = e^{-\frac{1}{3}} \] for the interval $ 0 \leq t \leq \pi $.

Step 2: Compute $ \frac{dx}{dt} $

To find the area, we first compute the derivative of $ x $ with respect to $ t $: \[ \frac{dx}{dt} = 6 \cos(t) \]

Step 3: Set Up the Area Integral

The area $ A $ enclosed by the curve and the y-axis can be expressed as: \[ A = \int_{0}^{\pi} y \frac{dx}{dt} \, dt \] Substituting $ y $ and $ \frac{dx}{dt} $: \[ A = \int_{0}^{\pi} e^{-\frac{1}{3}} \cdot 6 \cos(t) \, dt \]

Step 4: Evaluate the Integral

The integral simplifies to: \[ A = 6 e^{-\frac{1}{3}} \int_{0}^{\pi} \cos(t) \, dt \] The integral of $ \cos(t) $ over the interval $ [0, \pi] $ is: \[ \int_{0}^{\pi} \cos(t) \, dt = [\sin(t)]_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 \] Thus, the area becomes: \[ A = 6 e^{-\frac{1}{3}} \cdot 0 = 0 \]