# Maximize the integral of an implicit relation with two parameters

Suppose we have an equation: $${(r\cos{(t)}+u})^{2}+{(-(r\sin{(t)}+v)^3+1)}^{\frac{2}{3}}=1$$

Where $x=r\cos{t}$ and $y=r\sin{t}$

The graph of this implicit relation has two regions, where one part is above the x-axis, and the other part is below the x-axis.

Suppose we're are trying to find the area of the relation above the x-axis, between the x-values ${0}\le{x}\le{2\pi}$. How can one solve for the values of u and v that can give the highest area for this equation that's is above the x-axis, so that u and v as point (u,v) is still inside the implicit relation $$(1-x^2)^3=(1-y^3)$$.

Note: The blue stripes should stop at $2\pi$, and if the equation has regions that are "UNDEFINED", try to ignore it.

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Here I made some edits. Hopefully, this could help.

( 2014-12-17 11:52:53 -0500 )edit

Is there anyway I can solve this problem?

( 2014-12-21 15:48:10 -0500 )edit

I doubt my question became clearer. It's been months. If u and v has values that make the function discontinuous isn't part of the answer I'm looking for.

( 2015-01-08 18:38:12 -0500 )edit

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# Maximize the integral of an implicit function with two parameters

Title. Choosing a good title can help to attract interest to your question. Your original title, "How to use sage to solve this problem?", says nothing about the question you want to solve. Actually, remembering we are on ask-sage, it's sort of an empty title. I suggest changing the title of your question to "Maximize the integral of an implicit function with two parameters".

Notation. Choosing good notation can help geometric intuition.

The relation you want to study involves $u + y \cos x$ and $v + y \sin x$, so it would be more natural to use $t$ and $r$ instead of $x$ and $y$, and to use $x$ to denote $u + r \cos t$ and $y$ to denote $v + r \sin t$.

Reworking the relation to eliminate fractional exponents, we see that the question is about polar parametrizations with respect to shifted origins $(u, v)$ of the curve $C$ defined by $$(1-x^2)^3 = (1-y^3)^2.$$

The questions you are interested in are then:

• for which choices of $(u, v)$ does the relation define $r$ as a function of $t$?
• among those, which one gives the largest average for $r(t)$?

One way you can use Sage to get insight into this question is by plotting the curve $C$.

sage: f(x,y) = (1-x^2)^3 - (1-y^3)^2
sage: implicit_plot(f,(-1.02,1.02),(-0.02,1.3))


You see $C$ is a closed curve enclosing a nonconvex domain.

Appropriate choices of an origin $(u, v)$ for a polar parametrization $r(t)$ are those with respect to which the domain enclosed by $C$ is star-shaped.

Given the look of $C$ at $(0,0)$, $(1,1)$, $(-1,1)$, it seems that $(u, v) = (0,1)$ is in fact the only appropriate choice, and that no other choice will let you parametrize $C$ by a continuous function $r(t)$.

Given that, the maximization problem becomes trivial...

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@Slelievere How did you figure this out by using sage software. Can you show me?

( 2015-12-24 11:51:14 -0500 )edit