Problem 30 - Essential Calculus

Question

Use the disk or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about each given line. \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3}, \quad a>0\) (hypocycloid) (a) the \(x\) -axis (b) the \(y\) -axis

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by the hypocycloid around the x-axis and the y-axis can be calculated using the disk method, and the integral in the volume formula needs to be solved by the substitution method.

Step 1: Revolving around the x-axis

The given graph of a hypocycloid is symmetrical with respect to both the x and y axes. By looking at only the upper half of the curve from y = 0 to y = a and rotating it around the x-axis, the solid formed is a complete hypocycloid. Using the disk method for rotation around the x-axis, the general formula for volume is \(V=\pi \int_{a}^{b}[f(x)]^{2} dx\). In this case, the function f(x) can be derived from the equation of the hypocycloid as \(f(x)=\left(a^{\frac{2}{3}}-x^{\frac{2}{3}}\right)^{\frac{3}{2}}\). The limits of integration would be from 0 to \(a^{2/3}\). This gives the volume V.

Step 2: Calculate the volume for the x-axis

Substitute \(f(x)\) into the formula to obtain \(V=\pi \int_{0}^{a^{\frac{2}{3}}}\left[\left(a^{\frac{2}{3}}-x^{\frac{2}{3}}\right)^{\frac{3}{2}}\right]^{2} dx\). To solve the integral, one could substitute \(u = a^{2/3}-x^{2/3}\), then solve. Next, go back to x variable by substituting \(u\) back.

Step 3: Revolving around the y-axis

When revolving around the y-axis, consider the right side of the curve from x=0 to x=\(a^{2/3}\). Again, use the disk method. Here, the general formula for volume would be \(V=\pi \int_{c}^{d}[g(y)]^{2} dy\), where \(g(y)\) is derived from the equation of the hypocycloid to be \(g(y) = \left(a^{2/3}-y^{2/3}\right)^{3/2}\), and the limits of integration would go from 0 to \(a^{2/3}\).

Step 4: Calculate the volume for the y-axis

Substitute \(g(y)\) into the volume formula to calculate the integral, \(V=\pi \int_{0}^{a^{\frac{2/3}}}\left[\left(a^{2 / 3}-y^{2 / 3}\right)^{\frac{3}{2}}\right]^{2} dy\). This encountered integral can be solved with a similar substitution method as in Step 2. Substitute \(v = a^{2/3}-y^{2/3}\), then solve and replace \(v\) back.