To find the centre of gravity of the volume generated by revolution of the cardioid $r=a(1+\cos \phi)$ about the $x$-axis.

Question

Answer to the above question is $\left(\frac {4a} 5,0\right)$. Also, please explain if there's something wrong with the question since I don't get whether to find CG of volume or surface generated by the given curve.

Answer 1

Given the result they are asking for the centroid of the volume. (I therefore have edited the title of the question.)

We shall use the following parametrization of the upper half of the cardioid: $$\left.\eqalign{x(\phi)&=a(1+\cos\phi)\cos\phi\cr y(\phi)&=a(1+\cos\phi)\sin\phi\cr}\right\}\qquad(0\leq\phi\leq\pi)\ .$$

It is obvious that the centroid lies on the $x$-axis. We dissect the volume into cylindrical plates of radius $y(\phi)$ and thickness $|x'(\phi)|\>d\phi$. The total volume $V$ then is $$-\int_0^\pi \pi y^2(\phi)\>x'(\phi)\>d\phi={8\pi\over3}\,a^3\ .$$ Writing an overall minus sign here makes everything come out right: For $0\leq\phi\leq{2\pi\over3}$ one has $$x'(\phi)=-(1+2\cos\phi)\sin\phi\leq0\ ,$$ and for ${2\pi\over3}\leq\phi\leq\pi$ one has $x'(\phi)\geq0$. But the corresponding volume part (the inwards peak) has to be deducted.

In a second step we compute the total moment of all these cylinders in the $x$-direction. This leads to the integral $$-\int_0^\pi x(\phi)\>\pi y^2(\phi)\>x'(\phi)\>d\phi={32\pi\over15}\,a^4\ .$$ The $x$-coordinate $\xi$ of the centroid therefore is the quotient $$\xi={{32\pi\over15}a^4\over{8\pi\over3}a^3}={4a\over5}\ .$$

Answer 2

Due to symmetry, the CG lies on the $x$-axis, whose coordinate is

$$ x_c = \frac{\int_V x dxdydz}{\int_V dxdydz} = \frac{2\pi\int_0^\pi\int_0^{a(1+\cos\phi)} r\cos\phi r^2\sin\phi d\phi}{2\pi\int_0^\pi\int_0^{a(1+\cos\phi)} r^2\sin\phi d\phi} =\frac{\frac{16}{15}a^4}{\frac43a^3} =\frac45a $$