Given $\gamma(t) = (t-t^2,1-t^3), t \in [0,1]$, I want to calculate the area of the revolution of the solid when $\gamma$ rotates on the $y$-axis.
Now, given I can call $x = x(t)$ and $y = y(t)$, I have the following result:
$$ A = 2\pi\int_0^1 x(t)\sqrt{(x(t)')^2 + (y(t)')^2)}\, dt $$
I do all the calculations and I get: $$A = 2\pi\int_0^1 (t-t^2)\sqrt{1 -4t + 4t^2 + 9t^4}\,dt $$ But now I have no idea how to solve this. I tried splitting it and using substitution but I am getting nowhere.
Any tips?
You have the square root of a quartic with no repeated roots. The answer can only be written in terms of elliptic integrals – but I am going to derive such an answer anyway. All elliptic integrals and functions in this answer follow Mathematica/mpmath conventions.$\newcommand{sn}{\operatorname{sn}}\newcommand{cn}{\operatorname{cn}}$
I use Byrd and Friedman's Handbook of Elliptic Integrals for Engineers and Physicists. There are formulas in that book for integrals involving the square roots of general quartics with all complex roots – of which the one here, $9t^4+4t^2-4t+1$, is an example – but they are very unwieldy. So I am going to "soften up" the integral first by making a substitution that eliminates the cubic and linear terms of $P$.
Such a substitution is of the linear fractional type, $u=\frac{t+k}{lt+1}$, $t=\frac{k-u}{lu-1}$. When I put the $t$ expression into $P$ and solve for $k$ and $l$ that make the cubic and linear terms of the resulting expression in $u$ (after multiplying by $(lu-1)^4$) zero, I get one solution as $k=0,l=-1$. So let $u=\frac t{1-t}$: $$A=2\pi\int_0^\infty\frac{u\sqrt{10u^4-2u^2+1}}{(u+1)^6}\,du=2\pi B$$ $$B=\int_0^\infty\frac{u(10u^4-2u^2+1)}{(u+1)^6\sqrt{10u^4-2u^2+1}}\,du$$ So many powers of $u+1$ in the denominator would make it a pain if I used the tables now. Thus I perform a partial fraction decomposition of the rational function in $B$'s integrand… $$B=\int_0^\infty\left(\frac{10}{u+1}-\frac{50}{(u+1)^2}+\frac{98}{(u+1)^3}-\frac{94}{(u+1)^4}+\frac{45}{(u+1)^5}-\frac9{(u+1)^6}\right)\frac1{\sqrt{10u^4-2u^2+1}}\,du$$ …then reduce the powers using B&F 250.02: let $P$ be a quartic, $p$ a non-root of $P$ and $n>1$, then $$I_{-n,p}=\int_a^b\frac1{(t-p)^{-n}\sqrt P(t)}\,dt=\frac1{2(1-n)P(p)}\left( \left[\frac{2\sqrt{P(t)}}{(t-p)^{n-1}}\right]_a^b + \sum_{j=1}^4\frac{2m-2-j}{j!}P^{(j)}(p)I_{j-n,p}\right)$$ This reduces $I_{-n,p}$ to a rational linear combination of $1,I_{-1,p},I_{0,p},I_{1,p},I_{2,p}$, and in this case (with $a=0,b=\infty,P(u)=10u^4-2u^2+1$) $$B=\frac16 I_{-1,-1} - \frac{14}9 I_{0,-1} + \frac{898}{243} I_{1,-1} + \frac{449}{243}\left( - I_{2,-1} + \frac{\sqrt{10u^4-2u^2+1}}{10(u+1)} \right) - \frac{373}{1215} + \frac{77\sqrt{10}}{540}$$ Now let $J_n=\int_0^\infty\frac{t^n}{\sqrt{P(t)}}\,dt$. When rewritten in terms of $J_n$, $B$ has no $J_1$ term: $$B=\frac16 I_{-1,-1} + \frac{71}{243} J_0 + \frac{449}{243}\left( - J_2 + \frac{\sqrt{10u^4-2u^2+1}}{10(u+1)} \right) - \frac{373}{1215} + \frac{77\sqrt{10}}{540}$$ I have bracketed two terms together because they are infinite when standing alone but finite when combined. Now work out $J_2$ using B&F 225.09 and 361.61: $$J_2=\frac{10^{-3/4}}2((F(\varphi,m)-2E(\varphi,m)) + \frac{u\sqrt{10u^4-2u^2+1}}{10u^2+\sqrt{10}}$$ $$\varphi=2\tan^{-1}10^{1/4}u,m=\frac12+\frac1{2\sqrt{10}}$$ The sum of the algebraic part of $J_2$ and the other term in the bracket tends to $-\frac1{\sqrt{10}}$ as $u\to\infty$; $\varphi$ tends to $\pi$ in the same limit, so the integrals become complete. Thus the bracketed expression is $$(\cdot)=-10^{-3/4}(K(m)-2E(m))-\frac1{\sqrt{10}}$$ I can work out $J_0$ similarly using B&F 266.00; it's $10^{-1/4}K(m)$.
To work out $I_{-1,-1}$ use B&F 266.09. Letting $a=10^{-1/4}$: $$I_{-1,-1}=\frac a2\int_0^{2K(m)}\left(\frac{a\sn u}{1-\cn u}+1\right)^{-1}\,du=\frac a2\int_0^{2K(m)}\frac{-a\cn u}{2a+(a^2+1)\sn u}\,du+\frac a2\int_0^{2K(m)}\frac{a+\sn u}{2a+(a^2+1)\sn u}\,du$$ Of these last two integrals, the $\cn$-containing one evaluates to zero. This is because $\sn$ is even and $\cn$ is odd about $K(m)$, which the integration interval is symmetric over. The same properties allow us to halve the interval of the other integral. Hence $$I_{-1,-1}=\frac a2\int_0^{K(m)}\frac{1+(1/a)\sn u}{1+(a^2+1)/(2a)\sn u}\,du$$ Now I would apply B&F 361.69 here, but it seems incorrect and there is no errata for it. Using Mathematica (in Try It Online) I get $$I_{-1,-1}=\frac a2\left(cK(m)+(1-c)(Z + \Pi(n,m))\right)$$ $$c=\frac{2(10-\sqrt{10})}9,Z=\frac{\sqrt{440+161\sqrt{10}}}{27}\log(-3-\sqrt{10}),n=\frac12+\frac{11\sqrt{10}}{40}$$ Finally we have $$\boxed{A=2\pi(C_1K(m)+C_2E(m)+C_3(Z+\Pi(n,m))+C_4)}$$ $$C_1=10^{-1/4}\cdot\frac{580-247\sqrt{10}}{1215}\quad C_2=10^{-1/4}\cdot\frac{449\sqrt{10}}{1215}\quad C_3=10^{-1/4}\cdot\frac{2\sqrt{10}-11}{108}$$ $$C_4=-\frac{373}{1215}-\frac{41\sqrt{10}}{972}\quad m=\frac12+\frac1{2\sqrt{10}}\quad n=\frac12+\frac{11}{4\sqrt{10}}$$ $$Z=\frac{\sqrt{440+161\sqrt{10}}}{27}\log(-3-\sqrt{10})$$