A parametric curve is defined \(x = cos\left(\frac{\Pi t}{2}\right) , Y= sin\left(\frac{\Pi t}{2}\right)\) in the range of \(0\leq t\leq 1\) . It is rotated about X-axis by 360°. ‘What is the area of the surface generated?
2π
Understanding the Problem: Surface Area of Revolution
We are given a parametric curve defined by the equations \(x = \cos\left(\frac{\Pi t}{2}\right)\) and \(y = \sin\left(\frac{\Pi t}{2}\right)\) for the range \(0 \leq t \leq 1\). This curve is rotated completely (360°) about the X-axis. We need to find the area of the surface generated by this rotation, known as the surface area of revolution.
Analyzing the Parametric Curve
The given parametric equations \(x = \cos\left(\frac{\Pi t}{2}\right)\) and \(y = \sin\left(\frac{\Pi t}{2}\right)\) resemble the standard parameterization of a circle, \(x = \cos(\theta), y = \sin(\theta)\), where the parameter here is \(\theta = \frac{\Pi t}{2}\).
- When \(t = 0\), the parameter is \(\frac{\Pi \cdot 0}{2} = 0\). The point on the curve is \(x = \cos(0) = 1\), \(y = \sin(0) = 0\). This is the point (1, 0).
- When \(t = 1\), the parameter is \(\frac{\Pi \cdot 1}{2} = \frac{\Pi}{2}\). The point on the curve is \(x = \cos\left(\frac{\Pi}{2}\right) = 0\), \(y = \sin\left(\frac{\Pi}{2}\right) = 1\). This is the point (0, 1).
As \(t\) varies from 0 to 1, the parameter \(\frac{\Pi t}{2}\) varies from 0 to \(\frac{\Pi}{2}\). This means the curve traces out the part of the unit circle (\(x^2 + y^2 = 1\)) starting from (1, 0) and ending at (0, 1). This is a quarter circle in the first quadrant.
Formula for Surface Area of Revolution about the X-axis
For a parametric curve \(x = x(t), y = y(t)\) rotated about the X-axis from \(t=t_1\) to \(t=t_2\), the surface area of revolution (\(S\)) is given by the integral:
\(S = \int_{t_1}^{t_2} 2\Pi y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\)
The term \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\) represents the differential arc length, \(ds\).
Calculating the Necessary Derivatives
We need to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
- \(x = \cos\left(\frac{\Pi t}{2}\right)\)
\(\frac{dx}{dt} = \frac{d}{dt}\left[\cos\left(\frac{\Pi t}{2}\right)\right] = -\sin\left(\frac{\Pi t}{2}\right) \cdot \frac{d}{dt}\left(\frac{\Pi t}{2}\right) = -\frac{\Pi}{2}\sin\left(\frac{\Pi t}{2}\right)\) - \(y = \sin\left(\frac{\Pi t}{2}\right)\)
\(\frac{dy}{dt} = \frac{d}{dt}\left[\sin\left(\frac{\Pi t}{2}\right)\right] = \cos\left(\frac{\Pi t}{2}\right) \cdot \frac{d}{dt}\left(\frac{\Pi t}{2}\right) = \frac{\Pi}{2}\cos\left(\frac{\Pi t}{2}\right)\)
Calculating the Arc Length Differential Component
Now we find the term \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\):
\(\left(\frac{dx}{dt}\right)^2 = \left(-\frac{\Pi}{2}\sin\left(\frac{\Pi t}{2}\right)\right)^2 = \frac{\Pi^2}{4}\sin^2\left(\frac{\Pi t}{2}\right)\)
\(\left(\frac{dy}{dt}\right)^2 = \left(\frac{\Pi}{2}\cos\left(\frac{\Pi t}{2}\right)\right)^2 = \frac{\Pi^2}{4}\cos^2\left(\frac{\Pi t}{2}\right)\)
\(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \frac{\Pi^2}{4}\sin^2\left(\frac{\Pi t}{2}\right) + \frac{\Pi^2}{4}\cos^2\left(\frac{\Pi t}{2}\right)\)
\(= \frac{\Pi^2}{4}\left(\sin^2\left(\frac{\Pi t}{2}\right) + \cos^2\left(\frac{\Pi t}{2}\right)\right)\)
Using the trigonometric identity \(\sin^2(\theta) + \cos^2(\theta) = 1\), we get:
\(= \frac{\Pi^2}{4} \cdot 1 = \frac{\Pi^2}{4}\)
So, \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{\frac{\Pi^2}{4}} = \frac{\Pi}{2}\) (since \(\frac{\Pi}{2} > 0\)).
Setting up the Surface Area Integral
Now substitute \(y(t) = \sin\left(\frac{\Pi t}{2}\right)\), the arc length differential \(\frac{\Pi}{2}\), and the limits of integration \(t_1 = 0\), \(t_2 = 1\) into the surface area formula:
\(S = \int_{0}^{1} 2\Pi \left(\sin\left(\frac{\Pi t}{2}\right)\right) \left(\frac{\Pi}{2}\right) dt\)
\(S = \int_{0}^{1} \Pi^2 \sin\left(\frac{\Pi t}{2}\right) dt\)
Evaluating the Integral to Find the Surface Area
\(S = \Pi^2 \int_{0}^{1} \sin\left(\frac{\Pi t}{2}\right) dt\)
To evaluate this integral, we can use a substitution. Let \(u = \frac{\Pi t}{2}\). Then, \(du = \frac{\Pi}{2} dt\), which means \(dt = \frac{2}{\Pi} du\).
We also need to change the limits of integration according to the substitution:
- When \(t = 0\), \(u = \frac{\Pi \cdot 0}{2} = 0\).
- When \(t = 1\), \(u = \frac{\Pi \cdot 1}{2} = \frac{\Pi}{2}\).
The integral becomes:
\(S = \Pi^2 \int_{0}^{\frac{\Pi}{2}} \sin(u) \left(\frac{2}{\Pi} du\right)\)
\(S = \Pi^2 \cdot \frac{2}{\Pi} \int_{0}^{\frac{\Pi}{2}} \sin(u) du\)
\(S = 2\Pi \left[-\cos(u)\right]_{0}^{\frac{\Pi}{2}}\)
Now, apply the limits of integration:
\(S = 2\Pi \left(-\cos\left(\frac{\Pi}{2}\right) - (-\cos(0))\right)\)
We know that \(\cos\left(\frac{\Pi}{2}\right) = 0\) and \(\cos(0) = 1\).
\(S = 2\Pi \left(-0 - (-1)\right)\)
\(S = 2\Pi \left(0 + 1\right)\)
\(S = 2\Pi \cdot 1\)
\(S = 2\Pi\)
The surface area generated by rotating the parametric curve \(x = \cos\left(\frac{\Pi t}{2}\right), y = \sin\left(\frac{\Pi t}{2}\right)\) for \(0 \leq t \leq 1\) about the X-axis is \(2\Pi\).
| Step | Description | Result/Formula Used |
|---|---|---|
| 1 | Identify curve and range | Quarter circle \(0 \leq t \leq 1\) |
| 2 | Rotation axis | X-axis |
| 3 | Surface Area Formula (X-axis, Parametric) | \(S = \int_{t_1}^{t_2} 2\Pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\) |
| 4 | Calculate derivatives \(dx/dt, dy/dt\) | \(-\frac{\Pi}{2}\sin\left(\frac{\Pi t}{2}\right), \frac{\Pi}{2}\cos\left(\frac{\Pi t}{2}\right)\) |
| 5 | Calculate \(\sqrt{(dx/dt)^2 + (dy/dt)^2}\) | \(\frac{\Pi}{2}\) |
| 6 | Set up integral | \(\int_{0}^{1} 2\Pi \sin\left(\frac{\Pi t}{2}\right) \frac{\Pi}{2} dt = \int_{0}^{1} \Pi^2 \sin\left(\frac{\Pi t}{2}\right) dt\) |
| 7 | Evaluate integral | \(2\Pi\) |
Revision Table: Key Concepts
| Concept | Description | Formula (for rotation about X-axis) |
|---|---|---|
| Surface Area of Revolution | The area of the 3D shape created by rotating a curve around an axis. | Varies by curve type (parametric, function \(y=f(x)\), \(x=g(y)\)). |
| Parametric Curve | A curve defined by equations \(x=x(t), y=y(t)\), where \(t\) is a parameter. | Requires calculating \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). |
| Arc Length Differential (\(ds\)) | A small segment of the curve's length. | For parametric curves: \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\) |
| Rotation about X-axis | The curve spins around the horizontal axis. The radius of rotation for a point \((x,y)\) is \(|y|\). | Formula involves \(2\Pi y\) or \(2\Pi |y|\). |
Additional Information: Related Concepts
Understanding surface area of revolution involves several core calculus concepts:
- Integration: The total surface area is found by summing up infinitesimally small rings (generated by rotating \(ds\)) along the curve. This summation is done using a definite integral.
- Arc Length: The term \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\) is the arc length differential \(ds\). The integrand \(2\Pi y \, ds\) represents the area of a tiny band or frustum generated by rotating \(ds\) around the x-axis (with radius \(y\)).
- Other Formulas:
- For a curve \(y=f(x)\) rotated about the X-axis: \(S = \int_{a}^{b} 2\Pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx\)
- For a curve \(x=g(y)\) rotated about the X-axis: \(S = \int_{c}^{d} 2\Pi y \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} dy\)
- For a curve \(y=f(x)\) rotated about the Y-axis: \(S = \int_{a}^{b} 2\Pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx\)
- For a parametric curve rotated about the Y-axis: \(S = \int_{t_1}^{t_2} 2\Pi x(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\)
- Geometric Interpretation: Rotating the quarter circle in the first quadrant about the X-axis generates a hemisphere (half of a sphere). The surface area of a sphere with radius \(r\) is \(4\Pi r^2\). For a unit sphere (\(r=1\)), the area is \(4\Pi\). A hemisphere has half the surface area of a full sphere, which is \(\frac{1}{2} (4\Pi \cdot 1^2) = 2\Pi\). This matches our calculated result, providing a good check for our answer.
Important Questions from Definite Integrals
The Legendre polynomials P n(x), n = 0, 1, 2, ..., satisfying the orthogonailty condition \(\int_{{\rm{ - 1}}}^{\rm{1}} {{{\rm{P}}_{\rm{n}}}\left( {\rm{x}} \right){{\rm{P}}_{\rm{m}}}} \left( {\rm{x}} \right){\rm{dx}}\,{\rm{ = }}\,\frac{{\rm{2}}}{{{\rm{2n + 1}}}}{{\rm{\delta }}_{{\rm{nm}}}}\) on the interval [-1, +1], may be defined by the Rodrigues formula P n(x) = \(\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}{\rm{n!}}}}\frac{{{{\rm{d}}^{\rm{n}}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 1}}} \right)^{\rm{n}}}\) . The value of the definite integral \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \) is
- \(\rm \displaystyle\int_1^3 (e^{\log x} + 1) dx\) is equal to
if \(\displaystyle\int\dfrac{\sin x}{\sin (x-a)}dx=Ax+B\log |sin(x-a)|+ C\) where A, B and c are real constants then:
Which of the following is NOT a property of definite integral?
If \(\rm I_n = \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^n \theta \ d\theta \), then I8 + I6 equals:
