I have the cardioid $r=1+cos(t)$ for $0\leq{t}\leq{2\pi}$ and I want to calculate the surface of revolution of said curve. How can I calculate it?
The parematrization of the cardioid is:
$$x(t)=(1+cos(t))cos(t)$$
$$y(t)=(1+cos(t))sin(t)$$
and
$$\frac{dx}{dt}=\left(-2\cos\left(t\right)-1\right)\sin\left(t\right)$$
$$\frac{dy}{dt}=\cos\left(t\right)\left(\cos\left(t\right)+1\right)-\sin^2\left(t\right)
$$
To calculate the surface of revolution I know I can use the formula (since I want to revolve it around the x-axis) $$2\pi \int_{a}^{b} y(t)\sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2} dt$$
However, when I do that integral, the result is $0$ (using online calculators). Why is this wrong? Obviously the surface can't be 0, right? What's the correct answer?
Best Answer
First a tip: it is often easier to substitute back for $r$ in terms of $t$ after differentiating. We have $$\displaylines{ x=r\cos t\ ,\quad y=r\sin t\cr \frac{dx}{dt}=\frac{dr}{dt}\cos t-r\sin t\ ,\quad \frac{dy}{dt}=\frac{dr}{dt}\sin t+r\cos t\cr \eqalign{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2 &=\biggl(\frac{dr}{dt}\biggr)^2+r^2\cr &=\sin^2t+(1+\cos t)^2\cr &=2+2\cos t\cr &=4\cos^2\Bigl(\frac t2\Bigr)\cr}\cr \sqrt{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2} =2\Bigl|\cos\Bigl(\frac t2\Bigr)\Bigr|\ .\cr}$$ Note the absolute value - omitting this is often a reason why answers come out wrong. To find the area of the surface we have to integrate from $0$ to $\pi$, because rotating just the top half of the cardioid gives the whole surface: $$\eqalign{A &=2\pi \int_0^\pi y\sqrt{\bigg(\frac{dx}{dt}\bigg)^2 +\bigg(\frac{dy}{dt}\bigg)^2}\ dt\cr &=4\pi \int_0^\pi (1+\cos t)\sin t\Biggl|\cos\Bigl(\frac t2\Bigr)\biggr|\ dt\cr &=8\pi\int_0^{\pi/2} (1+\cos2u)\sin2u\,|\cos u|\,du\cr &=32\pi\int_0^{\pi/2} \cos^4u\sin u\,du\cr &=\frac{32\pi}5\ .\cr}$$