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I am faced with this question. I am asked to show that

[tex]

2(\sqrt{5}-2)\pi\leq\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA\leq\frac{\pi}{2}

[/tex]

Noting that the double integral is to be performed on region R which is bounded by the circle [tex]

x^2+y^2=1

[/tex]

From what I know, the double integral is a definite integral, then how come will it range from

[tex]

2(\sqrt{5}-2)\pi\leq

[/tex] to [tex]\frac{\pi}{2}[/tex]?

What I manage to do is by letting [tex]u=sin x[/tex] and [tex]v=sin y[/tex].

Therefore it follows that:

[tex]

\iint_{R}\frac{1}{\sqrt{4+u^2+v^2}}dA

[/tex]

And converting it into polar coodinates by letting [tex]u=rcos\theta[/tex] and [tex]v=rsin\theta[/tex] and [tex]dA=rdrd\theta[/tex], I have:

[tex]

\int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \frac{r}{\sqrt{4+r^2}}drd\theta

[/tex]

And then solving the above will give me [tex]2(\sqrt{5}-2)\pi[/tex]. However even then, this is a definite integral. How will I get a range of values? And even for that matter, how am I going to prove that

[tex]

2(\sqrt{5}-2)\pi\leq\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA

[/tex]

when all I have now is:

[tex]

2(\sqrt{5}-2)\pi=\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA

[/tex]

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# Double integral showing a range of values?

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