How do I calculate the surface area and volume of the solid obtained by rotating the Bernoulli lemniscate $$(x^2+y^2)^2=2a^2(x^2-y^2)$$ around the $x$-axis?
It is not like I'm lazy and asking for a ready solution, or completely helpless. I really tried to calculate this and I failed. That is why I'm asking for help :)
On
See the image (https://i.stack.imgur.com/a6cBL.png). Equation -: (x^2 + y^2)^2 = r^2*(x^2 - y^2). If the maximum distance from the center of the Lemniscate of Bernoulli (origin) to the end point on x-axis is 'r', then Volume obtained will be 0.45536*(r^3) and Surface Area will be 1.36608*(r^2). For proof, mail me at [email protected] .
The solution in the polar coordinates system where the lemniscate is given by the formula $r^{2}=2a^{2}\cos2\phi$.
Surface area can be obtained by using the formula $A=2\pi\int_a^br(\phi)\sin\phi\sqrt{r^{2}(\phi)+[dr(\phi)/d\phi]^{2}}d\phi$: $$A=2\pi a^{2}\sqrt{a}\int_a^b\sin\phi\sqrt{\cos2\phi +sin^{2}2\phi } d\phi$$ The integral above seems to have no analytical solutions.
The formula for the volume is pretty simple and can be obtained directly from the formula $V=\pi\int_a^b r^2(\phi)d\phi$ : $$V=2a^{2}\pi\int_a^b cos2\phi d\phi$$