- #1
caseyjay
- 20
- 0
Hi all,
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]
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]
Last edited: