Problem 49 - Essential Calculus

Question

Find the area of the zone of a sphere formed by revolving the graph of \(y=\sqrt{9-x^{2}}, 0 \leq x \leq 2\), about the \(y\) -axis.

Step-by-Step Solution

Answer

The area of the zone of the sphere formed by revolving the graph of the given function about the \(y\)-axis is \(23.382\pi\)

Step 1: Set up the integral

Plug the function \(f(x) = \sqrt{9-x^{2}}\) and the given bounds \(0 \leq x \leq 2\) into the integral formula. We obtain \(2\pi\int_{0}^{2}x\sqrt{9-x^{2}}dx\).

Step 2: Solve the Integral

This is a standard calculus problem at this point. Integrals of this type are generally reduced to a basic form using a trigonometric substitution. Here, we will use \(x = 3sin(\theta)\) because we have a sqrt(9 - x^2) which could be simplified using identity equation \(sin^2(\theta) + cos^2(\theta) = 1\). Now, replace \(x\) with \(3sin(\theta)\) in the integral and take the differential as well, which would give us \(dx = 3cos(\theta)d\theta\). This transforms our integral into a trigonometric integral: \(2\pi\int_{0}^{\arcsin(\frac{2}{3})}3sin(\theta).3\sqrt{9-9sin^{2}(\theta)}.3cos(\theta)d\theta\) which after simplification will become \(18\pi\int_{0}^{\arcsin(\frac{2}{3})}cos^{2}(\theta) d\theta\). With some further simplifying, we find that \(cos^{2}(\theta)\) can be written as \(\frac{1+cos(2\theta)}{2}\), so we end up integrating term-by-term: \(18\pi\int_{0}^{\arcsin(\frac{2}{3})}\frac{d\theta}{2} + 18\pi\int_{0}^{\arcsin(\frac{2}{3})}\frac{cos(2\theta)d\theta}{2}\)

Step 3: Calculation of the Integral

Now, we can simply solve these two integrals. The first one, \(\int d\theta\), is just \(\theta\), and the second one, \(\int cos(2\theta)d\theta\), is \(\frac{1}{2}sin(2\theta)\). We then evaluate these integrals at their bounds of 0 and \(\arcsin(\frac{2}{3})\), respectively. That gives us \(9\pi[\arcsin(\frac{2}{3}) - 0] + 9\pi[\frac{1}{2}sin(2\arcsin(\frac{2}{3})) - 0]\). After calculating this, we end up with the final surface area.