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$$int_{0}^{c} dy sqrt{frac{c-1/2y^2+1/3y^3}{1+2y}}$$ where c is a constant. This is coming from trying to find the area $$int_{U le c} dq_1dq_2$$ where $$U=frac{1}{2}(q_1^2+q_2^2)-frac{1}{3}q_2^3+q_1^2q_2$$ bounded by energy $c=U(q_1,q_2)$.

The Appell-Lauricella function is defined by the series
$$
F[{a,c};{b_1,b_2,dots,b_n};{x_1,x_2,ldots,x_n}]:=
$$
$$
=sum_{i_1,i_2,ldots,i_ngeq 0}frac{(a)_{i_1+i_2+ldots+i_n}(b_1)_{i_1}(b_2)_{i_2}ldots(b_n)_{i_n}}{(c)_{i_1+i_2+ldots+i_n}i_1!i_2!ldots i_n!}x_1^{i_1}x_2^{i_2}ldots x_n^{i_n},
$$
where $ngeq2$, $a,c,b_1,b_2,ldots,b_nintextbf{C}$ and $|x_1|<1,|x_2|<1,ldots,|x_n|<1$.

Then holds the following

THEOREM.
For $Re(c)>Re(a)>0$ and $|x_1|<1,|x_2|<1,ldots,|x_n|<1$, we have
$$
F[{a,c};{b_1,b_2,dots,b_n};{x_1,x_2,ldots,x_n}]=
$$
$$
=frac{Gamma(c)}{Gamma(a)Gamma(c-a)}int^{1}_{0}t^{a-1}(1-t)^{c-a-1}(1-x_1t)^{-b_1}(1-x_2t)^{-b_2}ldots (1-x_nt)^{-b_n}dt.
$$

Using the above theorem I will prove that

$$
int^{c}_{0}sqrt{frac{c-y^2/2+y^3/3}{1+2y}}dy=
frac{csqrt{4-l}}{2sqrt{6}}|l-1|times
$$
$$
times Fleft[{1,2};{frac{1}{2},-frac{1}{2},-frac{1}{2},-frac{1}{2}};{-2c,frac{2c}{l-1},frac{4c}{4-l-sqrt{3}sqrt{(4-l)l}},frac{4c}{4-l+sqrt{3}sqrt{(4-l)l}}}right],
$$
where $c=frac{1}{24}(4-9l+6l^2-l^3)$.

For to prove the above evaluation make the change of variable $yrightarrow -y$ to get
$$
int^{c}_{0}sqrt{frac{c-y^2/2+y^3/3}{2y+1}}dy=iint^{-c}_{0}sqrt{frac{y^2/2+y^3/3-c}{-2y+1}}dy,
$$
then $yrightarrow frac{1-w}{2}$ to get
$$
iint^{-c}_{0}sqrt{frac{y^2/2+y^3/3-c}{-2y+1}}dy=frac{sqrt{c}}{4sqrt{6}}int^{2c+1}_{1}sqrt{frac{24+1/c(w-4)(w-1)^2}{w}}dw.
$$
Now if $c=frac{1}{24}(4-9l+6l^2-l^3)$ we can write
$$
24+(-4+w)(-1+w)^2/c=frac{24(l-w)(9-6l+l^2-6w+lw+w^2)}{(l-4)(l-1)^2}.
$$
Hence we can write the last integral in the form of theorem and use it to get the result, which is the Appell-Lauricella function.

(Not an answer, just a comment that was too long).

You can prove that the value of the integral is $frac{c^2}{2sqrt{6}} + O(c)$ with the following algebraic simplifications. First note that the integral can be written as
$$ I = frac{1}{sqrt{6}}int_0^c sqrt{ (y-1)^2 + frac{6c-1}{2y+1}} dy. $$
It follows that
$$I > frac{1}{sqrt{6}} int_0^c (y-1) dy = frac{c(c-2)}{2sqrt{6}}.$$
Similarly, using the fact that $sqrt{a+b} < sqrt{a} + sqrt{b}$ (which does not quite hold in some regions of the domain but seems to be insignificant for large $c$), we have
$$ I < frac{1}{sqrt{6}} int_0^c (y-1) dy + frac{1}{sqrt{6}}int_0^c sqrt{frac{6c-1}{2y+1}} dy = frac{c^2}{2sqrt{6}} + O(c).$$
Thus we can conclude that $I = frac{c^2}{2sqrt{6}} + O(c).$ I think what could possibly help you is the following:

• If you want just a numerical answer, the function you are integrating is very smooth and convex so getting high precision values is doable.


• If you want a more accurate answer, you need to specify what regions of $c$ you are interested in. My answer holds for $c rightarrow infty$ but there are certainly more accurate answers for other cases such as $c << 1$.

HINT

The issue task is the task about the area under not convex figure.

Partially, for $C=0.135$ the graph is

for $C=frac16$ the graph is

and for $C=3.84$ the graph is

These figures show, that the proposed integral can't calculate the area ccorrectly, and it can be suitable to calculate the area in polar coordinates.

Let
$$q_1=rcos varphi, quad q_2=r sinvarphi,$$
then
$$U(r,varphi) = dfrac13r^3sin 3varphi+dfrac12r^2.tag1$$
Taking in account the properties of the sine function, it is sufficiently to consider $U(r,varphi)$ at the interval
$$varphiinleft(fracpi6,fracpi2right).$$
The bounds determines by the system of inequalities
begin{cases}
dfrac13r^3sin 3varphi+dfrac12r^2 > 0\
dfrac13r^3sin 3varphi+dfrac12r^2 < C.tag2
end{cases}
The first inequality has the solution
$$begin{cases}
rin(0,infty),quad text{if}quad varphiinleft(dfracpi6,dfracpi3right)\
rinleft(0,-dfrac3{2sin3varphi}right),quad text{if}quad varphiinleft(dfracpi3,dfracpi2right)
end{cases}tag3$$
Factor $dfrac4{Cr^3}$ allows to present the second inequality in the form of
$$dfrac{4}{r^3} - dfrac2{Cr} > dfrac{4sin3varphi}{3C},$$
or
$$4left(dfrac arright)^3-3dfrac{a}r > 2asin3varphi,tag4$$
where
$$a=sqrt{dfrac{3C}2}.tag5$$
If
$$2asin3varphi < 1,tag6$$ then can be used representation
$$cosleft(3arccosleft(dfrac arright)right) > 2asin3varphi.$$
Taking in account that
the inequality
$$cos(3arccos y) > x,quad xleq 1$$
has the solutions
$${small
yin
begin{cases}
[-1,1],quadtext{if}quad xin[-infty,-1)\
left[cosleft(dfrac13arccos xright),1right]bigcupleft[cosleft(dfrac{2pi}3+dfrac13arccos xright),cosleft(dfrac{2pi}3-dfrac13arccos xright)right],quadtext{if}quad xin[-1,1]
end{cases}}
$$
this means
$${small
rin
begin{cases}
[0,a],text{ if }pin[-infty,-1)\
left[a,dfrac a{cosleft(dfrac13arccos pright)}right] bigcupleft[dfrac a{cosleft(dfrac{2pi}3-dfrac13arccos pright)},dfrac a{cosleft(dfrac{2pi}3+dfrac13arccos pright)}right],text{ if } pin[-1,1],
end{cases}tag7}
$$
where
$$p = 2asin3varphi.tag8$$
If
$$2asin3varphi > 1,tag9$$ then can be used representation
$$coshleft(3cosh^{-1}left(dfrac arright)right) > 2asin3varphi,$$
$$ r < dfrac a{coshleft(dfrac13cosh^{-1}(2asin3varphi)right)},tag{10}$$
wherein
$$cosh^{-1}x = log(x+sqrt{x^2-1}),$$
$$coshleft(dfrac13cosh^{-1}xright)=dfrac12left(sqrt[3]{x+sqrt{x^2-1}}+dfrac1{sqrt[3]{x+sqrt{x^2-1}}}right).tag{11}$$
So for the conditions $(9)$
$$r < dfrac a{dfrac12left(sqrt[3]{p+sqrt{p^2-1}}+dfrac1{sqrt[3]{p+sqrt{p^2-1}}}right)}.tag{12}$$

At the same time, the area of figure in the polar coordinates equals to

$$S=6cdotdfrac12intlimits_{pi/6}^{pi/2}r^2(varphi),mathrm dvarphi.$$

This way looks more correct.