What is the Best Approach for Solving a Complicated Double Integral?

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SUMMARY

The forum discussion centers on solving the double integral \(\int_0^1\int_0^x \sqrt{4x^2-y^2} \, dy \, dx\). The user attempted various methods including a change of variables with \(t=4x^2-y^2\) and polar coordinates, but found them unhelpful. A key insight provided is that the inner integral resembles the standard integral \(\int \sqrt{a^2 - y^2} \, dy\), where \(a = 2x\), suggesting the use of trigonometric substitution as a viable approach.

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Hernaner28
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Homework Statement


I've got to calculate:

\displaystyle\int_0^1\displaystyle\int_0^x \sqrt{4x^2-y^2} dy dx


Homework Equations





The Attempt at a Solution



I've tried the change of variable:
\displaystyle t=4{{x}^{2}}-{{y}^{2}} but it doesn't get better. I've also tried polar coordinates but it is not convinient either. Do you know a convenient change? I've been trying to figure it out for a long time.

** I also tried changing the order of integration but no results.


Thanks!
 
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Hernaner28 said:

Homework Statement


I've got to calculate:

\displaystyle\int_0^1\displaystyle\int_0^x \sqrt{4x^2-y^2} dy dx

Homework Equations


The Attempt at a Solution



I've tried the change of variable:
\displaystyle t=4{{x}^{2}}-{{y}^{2}} but it doesn't get better. I've also tried polar coordinates but it is not convinient either. Do you know a convenient change? I've been trying to figure it out for a long time.

** I also tried changing the order of integration but no results.Thanks!

The y-integral (for any fixed value of x) is of the form
\int \sqrt{a^2 - y^2} \, dy,
where it happens that a = 2x. This is a standard integral that you must surely have seen before; if not, do a change of variables using a trigonometric substitution.

RGV
 
Last edited:
Thanks! I'll try!
 

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