Question

The surface area of the solid generated by revolving the curve \(x = e^t \cos t, y = e^t \sin t\) about y-axis for \(0 \le t \le \pi/2\) is

• A. \( 2\pi e^\pi (2-\sqrt{2}) \) sq. unit
• B. \( \frac{2\sqrt{2}}{5}\pi(e^\pi - 2) \) sq. unit
• C. \( \frac{2\sqrt{2}}{5}\pi(e^\pi + \pi) \) sq unit
• D. \( \frac{2\sqrt{2}}{5}\pi(e^\pi - \pi) \) sq unit

Hint:

The curve \(x=ae^{kt}\cos(kt), y=ae^{kt}\sin(kt)\) is a logarithmic spiral. Calculating \(ds\) for such curves often results in a neat simplification. The integral of the form \(\int e^{ax}\cos(bx)dx\) is very common in physics and engineering problems, and memorizing the formula can save a lot of time compared to performing integration by parts twice.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the surface area of a solid of revolution. The curve is given in parametric form, and it is revolved around the y-axis. The formula for surface area in this case involves integrating \(2\pi x\) multiplied by the arc length element \(ds\).
Step 2: Key Formula or Approach:
The surface area \(S\) generated by revolving a parametric curve \(x(t), y(t)\) from \(t=a\) to \(t=b\) about the y-axis is: \[ S = \int_a^b 2\pi x(t) \, ds \] where the arc length element \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\).
Step 3: Detailed Explanation:
1. Find Derivatives:
\( x = e^t \cos t \implies \frac{dx}{dt} = e^t\cos t - e^t\sin t = e^t(\cos t - \sin t) \)
\( y = e^t \sin t \implies \frac{dy}{dt} = e^t\sin t + e^t\cos t = e^t(\sin t + \cos t) \)
2. Find the Arc Length Element \(ds\):
\[ \left(\frac{dx}{dt}\right)^2 = e^{2t}(\cos^2 t - 2\sin t\cos t + \sin^2 t) = e^{2t}(1 - \sin(2t)) \] \[ \left(\frac{dy}{dt}\right)^2 = e^{2t}(\sin^2 t + 2\sin t\cos t + \cos^2 t) = e^{2t}(1 + \sin(2t)) \] \[ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = e^{2t}(1 - \sin(2t) + 1 + \sin(2t)) = 2e^{2t} \] \[ ds = \sqrt{2e^{2t}} \, dt = \sqrt{2} e^t \, dt \] 3. Set up and Evaluate the Integral:
The integral for the surface area is: \[ S = \int_{0}^{\pi/2} 2\pi (e^t \cos t) (\sqrt{2} e^t \, dt) = 2\sqrt{2}\pi \int_{0}^{\pi/2} e^{2t} \cos t \, dt \] We use the standard integral formula: \( \int e^{at}\cos(bt) dt = \frac{e^{at}}{a^2+b^2}(a\cos(bt) + b\sin(bt)) \). Here, a=2 and b=1. \[ \int e^{2t} \cos t \, dt = \frac{e^{2t}}{2^2+1^2}(2\cos t + \sin t) = \frac{e^{2t}}{5}(2\cos t + \sin t) \] Now evaluate the definite integral: \[ \left[ \frac{e^{2t}}{5}(2\cos t + \sin t) \right]_{0}^{\pi/2} = \left(\frac{e^{\pi}}{5}(2\cos(\pi/2) + \sin(\pi/2))\right) - \left(\frac{e^{0}}{5}(2\cos(0) + \sin(0))\right) \] \[ = \left(\frac{e^{\pi}}{5}(0 + 1)\right) - \left(\frac{1}{5}(2 + 0)\right) = \frac{e^{\pi}}{5} - \frac{2}{5} = \frac{e^\pi - 2}{5} \] 4. Calculate the Final Surface Area:
\[ S = 2\sqrt{2}\pi \left( \frac{e^\pi - 2}{5} \right) = \frac{2\sqrt{2}\pi}{5}(e^\pi - 2) \] Step 4: Final Answer:
The surface area is \( \frac{2\sqrt{2}}{5}\pi(e^\pi - 2) \) sq. unit.