In the class we were given a task to find a volume of a figure of revolution. The figure is an astroid $x=a\cos^3{t}, y=a\sin^3{t}$ axis is $x=a$. And I thought that instead of doing integration (wich we were meant to do) I had an idea of calculating it as following:
first calculate area of the astroid (the only part with integration), second find volume as product of area and its path. In our case path is a length of the circle of area $a$. Is it correct idea?
Here is my calculation
$$a\int_0^{\pi/2}\cos^3tdt=a\frac{2}{3}$$ this is the fourth part of area so the full area is $a\frac{8}{3}$. And the volume is
$$a\frac{8}{3}2\pi a=\frac{16}{3}\pi a^2$$
Is it correct? I mean not the calculation but do we have right to do it this way? (Without taking "volume integral" and moving axis).
Best Answer
Your idea is correct, but your calculus of the area is wrong.
The part of the area that you want is: $$ A/4=\int _0^a y(x) dx $$ note that here the integration is with respect to $x$ and not $t$. You can use the parametric equations to find $dx= 3a\cos^2 t (-\sin t) dt$ and substute in the integral with care to the limits of integration, and you find for your area:
$$ A/4=3a^2 \int_{\pi/2}^0 \sin^3 t \cos^2 t(-\sin t)dt $$ calculating this integral (see here) you find for the entire area $A=3 \pi a^2/8$. So the volume is $V=2\pi a A=3\pi^2a^3/4$