Problem 50 - Single Variable Calculus

Question

Verify that the surface area of a sphere of radius \(r\) is \(S=4 \pi r^{2}\) by evaluating a definite integral. Hint: Generate this sphere by revolving the semicircle \(x^{2}+y^{2}=r^{2}\) where \(y \geq 0\), about the \(x\) -axis.

Step-by-Step Solution

Answer

To verify the surface area of a sphere of radius \(r\), we set up the equation for a semicircle: \(x^2+y^2=r^2\), with \(y\geq0\). We compute the length element \(ds\), which is equal to \(\frac{r}{\sqrt{r^2-x^2}} dx\). Next, we find the surface element, \(dS\), by revolving the semicircle around the x-axis which gives us a surface element of \(2\pi y \cdot ds = 2\pi r dx\). We set up an integral for the surface area from \(-r\) to \(r\) and evaluate the integral to obtain the surface area: \( S = 4\pi r^2\). This verifies the given formula.

Step 1: Set up the equation for the semicircle

The equation of the semicircle with radius \(r\) and \(y \geq 0\) is given by: \[x^2 + y^2 = r^2\]

Step 2: Compute the length element \(ds\) for the semicircle

To find the length element \(ds\), we need to find the derivative of \(y\) with respect to \(x\): \[y^2 = r^2 - x^2\] \[y = \sqrt{r^2 - x^2}\] Differentiate \(y\) with respect to \(x\): \[\frac{dy}{dx} = \frac{-x}{\sqrt{r^2 - x^2}}\] Now, compute the length element \(ds\) using the Pythagorean theorem: \[ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx\] Plug in the value of \(\frac{dy}{dx}\): \[ds = \sqrt{1 + \left(\frac{-x}{\sqrt{r^2 - x^2}}\right)^2} dx = \frac{r}{\sqrt{r^2-x^2}} dx\]

Step 3: Set up the integral for the surface area

To find the surface area, the integral will be the sum of all the surface elements generated by revolving the semicircle around the x-axis. The surface element, \(dS\), can be represented as the product of the length element, \(ds\), and the distance traveled in a circle around the x-axis, which is \(2\pi y\): \[dS = 2\pi y \cdot ds\] Plug in the values of \(y\) and \(ds\) from the previous steps: \[dS = 2\pi (\sqrt{r^2 - x^2}) \cdot \frac{r}{\sqrt{r^2-x^2}} dx = 2\pi r dx\] Now, set up the integral for the surface area, \(S\), over the range of \(x\)-values for the semicircle, from \(-r\) to \(r\): \[S = \int _{-r}^{r} 2\pi r dx\]

Step 4: Evaluate the integral

Now, evaluate the integral to obtain the surface area: \[S = 2\pi r \int _{-r}^{r} dx = 2\pi r \bigg[ x \bigg]_{-r}^{r}\] \[S = 2\pi r (r - (-r)) = 2\pi r (2r) = 4\pi r^2\] The surface area of a sphere of radius \(r\) is found to be \(S = 4\pi r^2\), which verifies the given formula.

Key Concepts

Definite Integral Calculus

Definite integral calculus is a central concept in mathematics that enables us to calculate the accumulation of quantities, such as areas under curves, total distance traveled, or in our case, the surface area of a structure obtained through revolution. Specifically, it is the evaluation of the integral of a function within set bounds or limits. For a function f(x), the definite integral from a to b is represented as \[\int_{a}^{b} f(x) dx\].
To compute a definite integral, we apply the Fundamental Theorem of Calculus, which basically says that if F(x) is the antiderivative of f(x), then the definite integral of f(x) from a to b is given by the difference F(b) - F(a).

When applying definite integral calculus for the sphere's surface area, we consider the semicircle as an infinitely small slice rotating around an axis. By integrating along the lengths that these slices travel during the rotation, from one endpoint of the diameter to the other, we accumulate the overall surface area. This approach is a stunning application of how calculus helps us find properties of geometrical shapes that would otherwise be very difficult to compute.

Sphere Surface Area Derivation

The sphere surface area derivation through calculus offers us a fascinating and elegant way to comprehend the area of a three-dimensional object using two-dimensional analytical methods. In the provided problem, the semicircle's arc length at any point along the x-axis serves as the basis for deriving the sphere's surface area. Smaller elements of this arc, denoted as ds, when revolved around the x-axis, generate tiny bands on the sphere's surface—these add up to the total surface area.

Employing definite integral calculus in steps 3 and 4 of the solution translates the physical motion of revolution into a mathematical process. Through the setup \[ S = \int_{-r}^{r} 2\pi r dx\] and its subsequent evaluation, we arrive at the classic formula for a sphere's surface area: \[ S = 4\pi r^2 \].
This method offers a remarkable example of how calculus principles provide a bridge from the conceptually abstract to the concretely quantitative—showcasing the inherent beauty of mathematics in describing the physical world.

Revolution of Semicircle

The revolution of a semicircle around an axis to form a sphere is a concrete visual that's helpful in understanding the geometric origin of the formula for a sphere's surface area. When you rotate a semicircle around its diameter, every point on the boundary traces out a complete circle. These circles vary in size, being smallest at the poles (where they shrink to a single point) and largest at the equator of the sphere.

Calculating the surface area of the sphere by revolving a semicircle involves considering these circular trajectories as elements contributing to the whole surface. In step 2 of the solution, we determine the length element ds of the semicircle, which is necessary to understand how much surface area each infinitesimal segment of the semicircle contributes when rotated 360 degrees around the x-axis. Then in step 3, the product of \[2\pi y \cdot ds \] gives us dS, an element of surface area. By integrating these over the domain of x, we essentially 'add up' these bands to find the entire surface of the sphere. This geometric approach provides a tangible illustration of how motion through space is captured and expressed in the language of calculus.