Parametric Curve Surface Area Calculation

To find the surface area generated by rotating a parametric curve about the X-axis, we use the formula for the surface area of revolution. The given parametric curve is defined by:

• \(x = \cos \left( {\frac{{\pi u}}{2}} \right)\)
• \(y = \sin \left( {\frac{{\pi u}}{2}} \right)\)

The range for the parameter \(u\) is \(0 \le u \le 1\). The rotation is about the X-axis by 360 degrees.

Formula for Surface Area of Revolution

For a parametric curve rotated about the X-axis, the surface area \(A\) is given by the integral:

$$A = \int_{u_1}^{u_2} 2\pi y \sqrt{\left(\frac{dx}{du}\right)^2 + \left(\frac{dy}{du}\right)^2} du$$

Here, \(u_1 = 0\) and \(u_2 = 1\).

Step-by-Step Derivation

1. Derivatives with respect to \(u\)

First, we need to find the derivatives of \(x\) and \(y\) with respect to \(u\):

• For \(x = \cos \left( {\frac{{\pi u}}{2}} \right)\): $$\frac{dx}{du} = -\sin \left( {\frac{{\pi u}}{2}} \right) \cdot \frac{d}{du}\left({\frac{{\pi u}}{2}}\right) = -\sin \left( {\frac{{\pi u}}{2}} \right) \cdot \frac{\pi}{2}$$
• For \(y = \sin \left( {\frac{{\pi u}}{2}} \right)\): $$\frac{dy}{du} = \cos \left( {\frac{{\pi u}}{2}} \right) \cdot \frac{d}{du}\left({\frac{{\pi u}}{2}}\right) = \cos \left( {\frac{{\pi u}}{2}} \right) \cdot \frac{\pi}{2}$$

2. Calculate \( \left(\frac{dx}{du}\right)^2 + \left(\frac{dy}{du}\right)^2 \)

Next, we square each derivative and add them:

• $$\left(\frac{dx}{du}\right)^2 = \left(-\frac{\pi}{2}\sin \left( {\frac{{\pi u}}{2}} \right)\right)^2 = \frac{\pi^2}{4}\sin^2 \left( {\frac{{\pi u}}{2}} \right)$$
• $$\left(\frac{dy}{du}\right)^2 = \left(\frac{\pi}{2}\cos \left( {\frac{{\pi u}}{2}} \right)\right)^2 = \frac{\pi^2}{4}\cos^2 \left( {\frac{{\pi u}}{2}} \right)$$

Adding these two expressions:

$$\left(\frac{dx}{du}\right)^2 + \left(\frac{dy}{du}\right)^2 = \frac{\pi^2}{4}\sin^2 \left( {\frac{{\pi u}}{2}} \right) + \frac{\pi^2}{4}\cos^2 \left( {\frac{{\pi u}}{2}} \right)$$

Factor out \({\frac{{\pi^2}}{4}}\):

$$= \frac{\pi^2}{4} \left( \sin^2 \left( {\frac{{\pi u}}{2}} \right) + \cos^2 \left( {\frac{{\pi u}}{2}} \right) \right)$$

Using the trigonometric identity \({\sin^2 \theta + \cos^2 \theta = 1}\):

$$= \frac{\pi^2}{4} \cdot 1 = \frac{\pi^2}{4}$$

3. Calculate \( \sqrt{\left(\frac{dx}{du}\right)^2 + \left(\frac{dy}{du}\right)^2} \)

Now, take the square root:

$$\sqrt{\left(\frac{dx}{du}\right)^2 + \left(\frac{dy}{du}\right)^2} = \sqrt{\frac{\pi^2}{4}} = \frac{\pi}{2}$$

Since the curve segment is in the range \(0 \le u \le 1\), \({\frac{{\pi u}}{2}}\) is in the range \(0 \le {\frac{{\pi u}}{2}} \le {\frac{{\pi}}{2}}\). In this range, both \(\sin\) and \(\cos\) terms are such that the square root is positive.

4. Set up and Evaluate the Integral for Surface Area

Substitute \(y\) and the calculated square root into the surface area formula:

$$A = \int_{0}^{1} 2\pi \left(\sin \left( {\frac{{\pi u}}{2}} \right)\right) \left(\frac{\pi}{2}\right) du$$

Simplify the integrand:

$$A = \int_{0}^{1} \pi^2 \sin \left( {\frac{{\pi u}}{2}} \right) du$$

Now, evaluate the definite integral:

$$A = \pi^2 \left[ -\frac{\cos \left( {\frac{{\pi u}}{2}} \right)}{\frac{\pi}{2}} \right]_{0}^{1}$$

$$A = \pi^2 \left[ -\frac{2}{\pi} \cos \left( {\frac{{\pi u}}{2}} \right) \right]_{0}^{1}$$

$$A = -2\pi \left[ \cos \left( {\frac{{\pi u}}{2}} \right) \right]_{0}^{1}$$

Apply the limits of integration:

$$A = -2\pi \left( \cos \left( {\frac{{\pi \cdot 1}}{2}} \right) - \cos \left( {\frac{{\pi \cdot 0}}{2}} \right) \right)$$

$$A = -2\pi \left( \cos \left( {\frac{\pi}{2}} \right) - \cos(0) \right)$$

We know that \({\cos \left( {\frac{\pi}{2}} \right) = 0}\) and \({\cos(0) = 1}\):

$$A = -2\pi \left( 0 - 1 \right)$$

$$A = -2\pi \left( -1 \right)$$

$$A = 2\pi$$

Conclusion

The area of the surface generated by rotating the parametric curve about the X-axis is \(2\pi\).

The final answer is 2π.