Question

Let \(C\) be the curve \(x=f(t)\), \(y=g(t),\) for \(a \leq t \leq b,\) where \(f^{\prime}\) and \(g^{\prime}\) are continuous on \([a, b]\) and C does not intersect itself, except possibly at its endpoints. If \(g\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(x\)-axis is $$S=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$. Likewise, if \(f\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(y\)-axis is $$S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$ (These results can be derived in a manner similar to the derivations given in Section 6.6 for surfaces of revolution generated by the curve \(y=f(x)\).) Use the parametric equations of a semicircle of radius \(1, x=\cos t\) \(y=\sin t,\) for \(0 \leq t \leq \pi,\) to verify that surface area of a unit sphere is \(4 \pi\).

Short answer

Question: Using the parametric equations of a semicircle with radius 1, show that the surface area of a unit sphere is 4π. Answer: After computing the derivatives of the parametric equations and substituting them into the formula of the surface of revolution, we have evaluated the integral which yielded the surface area of the unit sphere as \(4 \pi\).

Step-by-step solution

Identify the parametric equations and the interval

The given parametric equations are: $$ x = f(t) = \cos t, $$ $$ y = g(t) = \sin t, $$where \(0 \leq t \leq \pi\).

Find the derivatives \(f'(t)\) and \(g'(t)\)

Differentiate the parametric equations with respect to \(t\): $$ f'(t) = -\sin t, $$ $$ g'(t) = \cos t, $$

Substitute into the formula for the surface of revolution

Since the curve is being revolved around the x-axis, we will use the formula: $$ S = \int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t. $$Substitute the given expressions for \(g(t)\), \(f'(t)\), and \(g'(t)\): $$ S = \int_{0}^{\pi} 2 \pi \sin t \sqrt{(-\sin t)^{2}+(\cos t)^{2}} d t. $$

Simplify the integrand and evaluate the integral

Inside the square root, notice that: $$ (-\sin t)^{2}+(\cos t)^{2} = \sin^2 t + \cos^2 t = 1, $$so the integrand simplifies to \(2 \pi \sin t\): $$ S = \int_{0}^{\pi} 2 \pi \sin t dt. $$Now, evaluate the integral: $$ S = 2 \pi [-\cos t]_{0}^{\pi} = 2 \pi (\cos 0 - \cos \pi) = 4 \pi. $$ Hence, by using the parametric equations of a semicircle of radius 1, we have verified that the surface area of a unit sphere is indeed \(4 \pi\).

Key concepts

Surface of Revolution

The concept of a surface of revolution involves creating a 3-dimensional shape by rotating a curve around an axis. Imagine taking a curve, like the semicircle in this problem, and spinning it 360 degrees around the x or y-axis. The shape that emerges from this rotation is called a surface of revolution.
A common example is revolving a semicircle around its diameter, which forms a sphere. The beauty of mathematics is that we can calculate the surface area of these shapes using calculus. The area formula involves integrating along the path of the curve, considering how the curve stretches out into a surface as it rotates.
By revolving - a curve given by parametric equations, - around the x-axis, for instance, - we use a specific formula that involves those equations and their derivatives to find the precise surface area.

Parametric Equations

Parametric equations are a powerful way to represent curves. They describe both the x-coordinate and the y-coordinate as functions of a third parameter, usually denoted as 't'.
In our problem, the parametric equations for a semicircle are given as:- \(x = \cos t\)- \(y = \sin t\).
This represents the x and y positions of points on the semicircle as 't' varies from 0 to \(\pi\). One of the biggest benefits of parametric equations is that they can represent complex curves that may be difficult to describe with a single function of x or y.
When dealing with surfaces of revolution, parametric equations allow mathematicians to handle curves that don't fit neatly into the form \(y = f(x)\). Their flexibility is invaluable, especially when integrating to find areas or volumes.

Calculus Integration

Calculus integration is a method for finding total quantities, such as areas or volumes, by adding infinitesimally small pieces together. In the context of surfaces of revolution, integration helps calculate the total surface area generated by rotating a curve.
For instance, when using parametric equations like \(x = \cos t\) and \(y = \sin t\) for the semicircle, integration involves:- finding derivatives, \(f'(t)\) and \(g'(t)\), to understand how each component of the curve changes,- setting up an integral formula for area that incorporates these derivatives and the parametric equations.
The integration simplifies complex geometric problems into solvable mathematical equations. In this example, the integral \(\int_{0}^{\pi} 2 \pi \sin t dt\) evaluates to \(4 \pi\), confirming the surface area of a sphere. This shows calculus integration as a critical tool in analyzing and solving problems related to geometric properties of surfaces.