Surface Area Calculation for Parametric Curve Revolution

This problem requires calculating the surface area generated when a parametric curve is revolved around the y-axis. The curve is defined by the equations:

\( x(t) = e^t \cos t \)

\( y(t) = e^t \sin t \)

over the interval \( 0 \leq t \leq \frac{\pi}{2} \).

The formula for the surface area \( S \) generated by revolving a parametric curve \( x = x(t), y = y(t) \) about the y-axis is:

\[ S = \int_{a}^{b} 2\pi x \, ds \]

where \( ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \). We must ensure \( x \geq 0 \) on the interval. For \( 0 \leq t \leq \frac{\pi}{2} \), \( e^t > 0 \) and \( \cos t \geq 0 \), therefore \( x(t) = e^t \cos t \geq 0 \), satisfying the condition.

Derivatives of Parametric Equations

We start by finding the derivatives of \( x(t) \) and \( y(t) \) with respect to \( t \), using the product rule.

• Derivative of \( x(t) \):

  \( \frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) \) \( = (e^t)' \cos t + e^t (\cos t)' \) \( = e^t \cos t + e^t (-\sin t) \) \( = e^t (\cos t - \sin t) \)

• Derivative of \( y(t) \):

  \( \frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) \) \( = (e^t)' \sin t + e^t (\sin t)' \) \( = e^t \sin t + e^t (\cos t) \) \( = e^t (\sin t + \cos t) \)

Arc Length Element Calculation

Next, we calculate the square of the arc length element, \( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \).

\( \left(\frac{dx}{dt}\right)^2 = \left(e^t (\cos t - \sin t)\right)^2 = e^{2t} (\cos^2 t - 2\cos t \sin t + \sin^2 t) \) \( = e^{2t} (1 - 2\cos t \sin t) \)

\( \left(\frac{dy}{dt}\right)^2 = \left(e^t (\sin t + \cos t)\right)^2 = e^{2t} (\sin^2 t + 2\sin t \cos t + \cos^2 t) \) \( = e^{2t} (1 + 2\sin t \cos t) \)

Summing these squares:

\( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = e^{2t} (1 - 2\cos t \sin t) + e^{2t} (1 + 2\sin t \cos t) \) \( = e^{2t} (1 - \sin(2t) + 1 + \sin(2t)) \) \( = e^{2t} (2) = 2e^{2t} \)

The arc length element differential \( ds \) is found by taking the square root:

\( \frac{ds}{dt} = \sqrt{2e^{2t}} = \sqrt{2} e^t \) (Since \( e^t > 0 \))

Surface Area Integral Setup

Now, we set up the integral for the surface area \( S \) using the formula \( S = \int_{a}^{b} 2\pi x \, ds \). We substitute \( x(t) = e^t \cos t \) and \( \frac{ds}{dt} = \sqrt{2} e^t \) with the limits \( a=0 \) and \( b=\frac{\pi}{2} \).

\( S = \int_{0}^{\pi/2} 2\pi (e^t \cos t) (\sqrt{2} e^t) \, dt \) \( S = 2\pi \sqrt{2} \int_{0}^{\pi/2} e^{2t} \cos t \, dt \)

Integral Evaluation for Surface Area

We need to evaluate the integral \( \int e^{2t} \cos t \, dt \). We use the standard integration formula for \( \int e^{at} \cos(bt) \, dt \):

\[ \int e^{at} \cos(bt) \, dt = \frac{e^{at}}{a^2+b^2} (a \cos(bt) + b \sin(bt)) + C \]

Here, \( a=2 \) and \( b=1 \). Thus, the integral is:

\( \int e^{2t} \cos t \, dt = \frac{e^{2t}}{2^2+1^2} (2 \cos t + 1 \sin t) = \frac{e^{2t}}{5} (2 \cos t + \sin t) \)

Now, we evaluate the definite integral from \( t=0 \) to \( t=\frac{\pi}{2} \):

\( S = 2\pi \sqrt{2} \left[ \frac{e^{2t}}{5} (2 \cos t + \sin t) \right]_{0}^{\pi/2} \)

Evaluating the expression at the limits:

• At \( t = \frac{\pi}{2} \): \( \frac{e^{2(\pi/2)}}{5} (2 \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})) = \frac{e^{\pi}}{5} (2 \cdot 0 + 1) = \frac{e^{\pi}}{5} \)
• At \( t = 0 \): \( \frac{e^{2(0)}}{5} (2 \cos(0) + \sin(0)) = \frac{e^0}{5} (2 \cdot 1 + 0) = \frac{1}{5} (2) = \frac{2}{5} \)

Subtracting the value at the lower limit from the value at the upper limit:

\( S = 2\pi \sqrt{2} \left( \frac{e^{\pi}}{5} - \frac{2}{5} \right) \)

Final Surface Area Result

Simplifying the result gives the total surface area:

\( S = \frac{2\pi \sqrt{2}}{5} (e^{\pi} - 2) \) square units.