The problem asks to find the value of $a$ ($a>0$) such that the volume $V_x$ of the solid formed by rotating the region enclosed by the parabola $y = \frac{1}{2}x^2 - ax$ and the $x$-axis around the $x$-axis is equal to the volume $V_y$ of the solid formed by rotating the same region around the $y$-axis.

1. Problem Description

The problem asks to find the value of $a$ ($a>0$) such that the volume $V_x$ of the solid formed by rotating the region enclosed by the parabola $y = \frac{1}{2}x^2 - ax$ and the $x$-axis around the $x$-axis is equal to the volume $V_y$ of the solid formed by rotating the same region around the $y$-axis.

2. Solution Steps

First, we find the intersection points of the parabola $y = \frac{1}{2}x^2 - ax$ and the $x$-axis ($y=0$).

$$$$

\frac{1}{2}x^2 - ax = 0 \\

x(\frac{1}{2}x - a) = 0

$$$$

So, $x=0$ or $x=2a$. Since $a > 0$, the intersection points are $x=0$ and $x=2a$.

Next, we calculate the volume $V_x$ obtained by rotating the region around the $x$-axis.

$$$$

V_x = \pi \int_{0}^{2a} (\frac{1}{2}x^2 - ax)^2 dx \\

= \pi \int_{0}^{2a} (\frac{1}{4}x^4 - ax^3 + a^2x^2) dx \\

= \pi [\frac{1}{20}x^5 - \frac{a}{4}x^4 + \frac{a^2}{3}x^3]_{0}^{2a} \\

= \pi [\frac{1}{20}(2a)^5 - \frac{a}{4}(2a)^4 + \frac{a^2}{3}(2a)^3] \\

= \pi [\frac{32a^5}{20} - \frac{16a^5}{4} + \frac{8a^5}{3}] \\

= \pi [\frac{8a^5}{5} - 4a^5 + \frac{8a^5}{3}] \\

= \pi a^5 [\frac{24 - 60 + 40}{15}] \\

= \pi a^5 [\frac{4}{15}] \\

= \frac{4\pi a^5}{15}

$$$$

Now, we calculate the volume $V_y$ obtained by rotating the region around the $y$-axis. We use the shell method.

$$$$

V_y = 2\pi \int_{0}^{2a} x |\frac{1}{2}x^2 - ax| dx

$$$$

Since $\frac{1}{2}x^2 - ax \le 0$ for $0 \le x \le 2a$,

$$$$

V_y = 2\pi \int_{0}^{2a} x (ax - \frac{1}{2}x^2) dx \\

= 2\pi \int_{0}^{2a} (ax^2 - \frac{1}{2}x^3) dx \\

= 2\pi [\frac{a}{3}x^3 - \frac{1}{8}x^4]_{0}^{2a} \\

= 2\pi [\frac{a}{3}(2a)^3 - \frac{1}{8}(2a)^4] \\

= 2\pi [\frac{8a^4}{3} - \frac{16a^4}{8}] \\

= 2\pi [\frac{8a^4}{3} - 2a^4] \\

= 2\pi a^4 [\frac{8}{3} - 2] \\

= 2\pi a^4 [\frac{2}{3}] \\

= \frac{4\pi a^4}{3}

$$$$

We are given that $V_x = V_y$.

$$$$

\frac{4\pi a^5}{15} = \frac{4\pi a^4}{3} \\

\frac{a}{15} = \frac{1}{3} \\

a = \frac{15}{3} \\

a = 5

$$$$

3. Final Answer

5