Question

Let $C$ be the curve $x=f(t)$, $y=g(t),$ for $a\le t\le b,$ 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 $$S={\int }_a^{b}2\pi g(t)\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}dt$$. Likewise, if $f$ is nonnegative on $[a,b],$ then the area of the surface obtained by revolving C about the $y$-axis is $$S={\int }_a^{b}2\pi f(t)\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}dt$$ (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 $y=f(x)$.) Consider the curve $x=3\cos t,y=3\sin t+4,$ for $0\le t\le 2\pi$ a. Describe the curve. b. If the curve is revolved about the $x$ -axis, describe the shape of the surface of revolution and find the area of the surface.

Short answer

Answer: The area of the surface of revolution is $144{\pi }^2+18\pi$.

Step-by-step solution

Describe the curve

First, we need to describe the given curve. To do this, we can utilize the fact that the curve's $x$ and $y$ coordinates can be expressed as functions of $t$, $x=3\cos (t)$ and $y=3\sin (t)+4$. Since the $x$ and $y$ coordinates are trigonometric functions, we can anticipate that the curve will be a type of circle or ellipse. Observe that the coefficients for $\cos (t)$ and $\sin (t)$ are the same, so the curve will be a circle with a radius of 3 units centered at $(0,4)$ on the Cartesian plane.

Revolve the curve about the x-axis and describe the surface

Next, we need to revolve the curve around the $x$-axis. The circle that is described by the parametric functions will generate a torus (also known as a doughnut shape) when revolved about the $x$-axis. The torus will have a center circle of radius 4 units and an outer circle of radius 7 units, with the distance between these circles being 3 units.

Calculate the area of the surface of revolution

Now, we'll calculate the area of the surface obtained by revolving the curve around the $x$-axis. We will use the formula provided in the exercise: $$S={\int }_a^{b}2\pi g(t)\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}dt$$ First, find the derivatives of the parametric functions of the curve: $$f^{\prime }(t)=\frac{d}{dt}(3\cos (t))=-3\sin (t)$$ $$g^{\prime }(t)=\frac{d}{dt}(3\sin (t)+4)=3\cos (t)$$ Substitute these into the formula: $$S={\int }_0^{2\pi }2\pi (3\sin (t)+4)\sqrt{(-3\sin (t){)}^2+(3\cos (t){)}^2}dt$$ This simplifies to: $$S={\int }_0^{2\pi }6\pi (3\sin (t)+4)dt$$ Integrating and evaluating over the range $0\le t\le 2\pi$: $$S=6\pi {[-3\cos (t)+4t]}_0^{2\pi }=6\pi (-3\cos (2\pi )+4(2\pi )-(-3\cos (0)+4(0)))$$ $$S=6\pi (-3+8\pi )=6\pi (8\pi +3)=144{\pi }^2+18\pi$$ Hence, the area of the surface generated by revolving the curve around the $x$-axis is $144{\pi }^2+18\pi$.

Key concepts

Parametric Equations

Parametric equations provide a powerful way to represent curves in a plane. Instead of describing a curve with a single equation involving just two variables like $x$ and $y$, we express them as separate functions of a third variable, usually $t$. This third variable, often called the parameter, allows for more flexible and dynamic curve representations.

For example, the curve given by the parametric equations $x=3\cos (t)$ and $y=3\sin (t)+4$ is a circle. Here, $t$ ranges from 0 to $2\pi$, which corresponds to making a full loop around the circle. The trigonometric functions $\cos (t)$ and $\sin (t)$ ensure that the values of $x$ and $y$ trace a path with circular symmetry.

• In this case, the curve is centered at $(0,4)$ on the Cartesian plane.
• It has a radius of 3, derived from the coefficients of $\cos (t)$ and $\sin (t)$.
• The parametric form ensures that as $t$ changes, we smoothly trace the entire circle.

Surface Area Calculation

When a curve is revolved around an axis, it generates a three-dimensional shape. To find the surface area of such a shape, we use specific calculus techniques. Surface area calculation for surfaces of revolution involves integrating the product of the circumference of small circles formed by the revolution and an infinitesimal arc length of the curve.

For a curve given in parametric form, the surface area $S$ obtained by revolving the curve about the x-axis is computed using:
$$S={\int }_a^{b}2\pi g(t)\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}dt$$
This equation integrates over the interval $[a,b]$, effectively summing up an infinite number of infinitesimally thin rings around the axis.

• The term $2\pi g(t)$ represents the circumference of these rings, based on their distance from the x-axis.
• $\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}$ gives the surface's infinitesimal arc length, derived from the derivatives of the parametric equations.

This integration results in the total surface area created by the revolution. It's particularly useful for finding areas of shapes like the torus (doughnut shape), generated when the given circle trials in this exercise is revolved around the x-axis.

Integration in Calculus

Integration is a fundamental concept within calculus, often used to compute areas, volumes, and other quantities that are summed over continuous ranges. In this context, it involves summing up infinitely small pieces to find a total value.

In our exercise, the integration process lets us find the surface area created by revolving our parameterized curve about an axis. Here's how it unfolds:

• First, derivatives $f^{\prime }(t)=-3\sin (t)$ and $g^{\prime }(t)=3\cos (t)$ are calculated. These help specify the rate at which the position changes along the parameter $t$.
• The formula $\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}$ then evaluates the infinitesimal arc length of the curve segment, crucial for integrating these segments over their continuous length.
• Finally, the definite integral ${\int }_0^{2\pi }6\pi (3\sin (t)+4)dt$ computes the sum of all surface area elements over the entire interval from $0$ to $2\pi$.

The calculations lead us to an evaluation of the integral as $S=144{\pi }^2+18\pi$. Each step in integration is key to turning abstract motion and rotation into a concrete surface area result, as seen in our surface of revolution problem.