Find the volume of astroid $x(t)=a\cos ^3t, y(t)=a\sin^3t$ which is revolved about $y$ - axis. Apply The cylindrical shell method.
My solution: The formula for cylindrical shell method is the following: $V=\int \limits_{a}^{b}2\pi xf(x)dx$. This astroid intersects axises at points $(\pm a, 0)$ and $(0,\pm a)$. The integral $\int \limits_{\frac{\pi}{2}}^{0}2\pi x(t)y(t)dx(t)$ is the volume of astroid (located in first quadrant) which is revolved about $y$ - axis. So the total area will be twice of above integral $$2\int \limits_{\frac{\pi}{2}}^{0}2\pi x(t)y(t)dx(t)=-4\pi \int \limits_{0}^{\frac{\pi}{2}}a\cos^3t \ a\sin^3t\ 3a\cos ^2t (-\sin t)dt=12\pi \int \limits_{0}^{\frac{\pi}{2}} \cos^5 t \sin ^4t dt=$$$$=12\pi B(3, \frac{5}{2})=12\pi \dfrac{\Gamma(3)\Gamma(5/2)}{\Gamma(11/2)}=12\pi \dfrac{\Gamma(3)\Gamma(5/2)}{\frac{9 \cdot 7 \cdot 5}{2^3}\Gamma(5/2)}=\dfrac{12\pi \cdot 2^4}{9\cdot 7 \cdot 5}=\dfrac{64 \pi}{105} $$ However the real answer is half of mine, namely $\dfrac{32 \pi}{105}$.
Where did I do mistake? I think that my solution is correct :/
Here is perhaps an easier way to do this. The leg of the astroid in the first quadrant is given by
$$y=(1-x^{2/3})^{3/2},\quad x\in[0,1]$$
Then, taking advantage of $x$-axis symmetry,
$$ \begin{align} V &=2\cdot2\pi\int_0^1 xy~dx\\ &=4\pi\int_0^1 x (1-x^{2/3})^{3/2}~dx\\ &=4\pi\frac{3}{2}\int_0^1 u^2(1-u)^{3/2}~du\\ &=6\pi~\text{B}(3,5/2)\\ &=6\pi~\frac{\Gamma(3)\Gamma(5/2)}{\Gamma(11/2)} \end{align} $$
The figure below shows the astroid of revolution. With this solution you can find the volume of all manner of curves $y=(1-x^q)^p,\quad x\in[0,1]$. These curves are part of the family of superconics that I have described previously here.