Question

Find the area of the zone of a sphere formed by revolving the graph of \(y=\sqrt{r^{2}-x^{2}}, 0 \leq x \leq a,\) about the \(y\) -axis. Assume that \(a

Short answer

The area of the zone of the sphere is given by \(A = 2\pi r[r\sin^{-1}\frac{a}{r}-a\sqrt{r^{2}-a^{2}}]\)

Step-by-step solution

Understanding the exercise

In this step, identify that the given equation represents a sphere and the task is to find the area of the zone of the sphere formed by revolving the given graph about the y-axis. This zone is the section of the sphere from x=0 to x=a. The restriction \(a<r\) signifies that we're dealing with a segment of the sphere, not the whole sphere.

Formulating the radius of the sphere

The radius of the sphere is given by \(r\). The given equation is the formula for a circle or sphere, where r is the radius of the sphere.

Derivation of the surface area formula

The formula for the surface area of a sphere is given by \(A = 4\pi r^{2}\). But in this case, when the sphere is revolved around the y-axis, we get a differential surface area at given x which can be obtained by \(dA = 2\pi x ds\), where \(ds\) is the arc length, which can be derived from the Pythagorean theorem as \(ds=\sqrt{dx^{2} + dy^{2}}\). Substituting the value of \(y\) will yield \(ds = dx\sqrt{1+y'^{2}}\).

Calculating the derivative of y and substituting it into ds

The derivative of \(y\) can be calculated using the power rule of differentiation. Therefore, \(y'=-\frac{x}{\sqrt{r^{2}-x^{2}}}\). Substituting the value of \(y'\) in the equation for ds gives \(ds=dx\sqrt{1+\left(\frac{x^{2}}{r^{2}-x^{2}}\right)}\). This simplifies to \(ds = \frac{r}{\sqrt{r^{2}-x^{2}}}dx\).

Substituting the value of ds into the formula of dA

Now that we have obtained \(ds\), substitute this value into the formula of \(dA\) to get \(dA = 2\pi x \cdot \frac{r}{\sqrt{r^{2}-x^{2}}} dx\).

Finding the total area

The total area, denoted by \(A\), is the integral of the differential area from \(x=0\) to \(x=a\). Therefore, \(A = \int_{0}^{a} dA = 2\pi r\int_{0}^{a} \frac{x}{\sqrt{r^{2}-x^{2}}}dx\). By using the substitution method let \(u = r^{2}-x^{2}\), the integral can be solved to yield \(A = 2\pi r[r\sin^{-1}\frac{a}{r}-a\sqrt{r^{2}-a^{2}}]\).

Consideration of given condition

Put a<r into the result. The solution above works for all \(a<=r\), but in this exercise, a is strictly less than r, which has to be noted.

Key concepts

Integral of Spherical Zones

When solving problems involving the surface area of a sphere, or sections known as spherical zones, we use integral calculus. A spherical zone is the portion of the sphere’s surface lying between two parallel planes that cut the sphere. Imagine peeling that particular 'strip' of the skin from the sphere – that's the zone we're referring to. In calculus, we calculate the area of this zone by integrating the area of infinitely small bands or strips that make up the zone over the interval we are interested in.

For example, in our exercise, we were tasked with finding a specific spherical zone created when a semicircle defined by the function \(y=\frac{r^{2} - x^{2}}{r}\) is revolved around the y-axis. The integral is set up over the x-interval [0, a], which represents the horizontal slice of the sphere under consideration. This approach breaks down the complex curved surface into calculable elements, which when summed (integrated) provide the total area of the zone.

Differentiation and Integration in Calculus

Differentiation and integration are two foundational concepts in calculus, acting as inverse processes. Differentiation is the process of finding the derivative, which gives the rate at which one quantity changes with respect to another. For instance, in our sphere surface area problem, differentiation was used to determine the rate at which the function \(y\) changes with respect to \(x\), given as \(y'=-\frac{x}{\frac{r^{2} - x^{2}}{r}}\).

Integration, on the other hand, is the process of finding the integral, representing the accumulation of quantities, such as areas under curves. After differentiating, we integrated the differential surface area formula to find the total area of the spherical zone. Integration solves for the whole when only parts are known, by summing an infinite number of infinitesimally small parts, as we did with the spherical strips.

Pythagorean Theorem in Calculus

The Pythagorean theorem is a staple in geometry that relates the sides of a right triangle. However, it can also appear in calculus problems, especially when dealing with curved surfaces or lengths of curves. Essentially, it helps in finding the length of the hypotenuse in the differential element we consider.

In the context of our exercise, when finding the arc length \(ds\), we are effectively looking at the hypotenuse of an infinitesimally small right triangle formed by \(dx\) and \(dy\). The formula \(ds=\frac{r}{\frac{r^{2} - x^{2}}{r}}dx\) stems from the Pythagorean theorem applied to the differentials. This allows us to account for the fact that the surface we are dealing with is not flat, but curved, and requires an extension of the simple concept of the Pythagorean theorem into the realm of calculus to properly measure the distance along a curve.

Arc Length Formula

The arc length formula is employed to calculate the distance along a curved path, which is crucial when dealing with objects like spheres. Mathematically, the arc length \(ds\) of a curve can be expressed using square roots and differentials: \(ds = \frac{r}{\frac{r^{2} - x^{2}}{r}}dx\). This takes into account the true path along the curve, not just the straight distance between two points.

The formula is derived from the Pythagorean theorem as previously discussed, and when you're dealing with a sphere, this distance becomes an element of the surface area. To obtain the surface area of a spherical zone, we integrate this arc length over the desired interval. As the sphere's curvature changes with latitude, the arc length changes accordingly, making the integral crucial for accurate calculations.