# Double integral showing a range of values?

1. Apr 9, 2009

### caseyjay

Hi all,

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

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

Noting that the double integral is to be performed on region R which is bounded by the circle $$x^2+y^2=1$$

From what I know, the double integral is a definite integral, then how come will it range from
$$2(\sqrt{5}-2)\pi\leq$$ to $$\frac{\pi}{2}$$?

What I manage to do is by letting $$u=sin x$$ and $$v=sin y$$.

Therefore it follows that:
$$\iint_{R}\frac{1}{\sqrt{4+u^2+v^2}}dA$$

And converting it into polar coodinates by letting $$u=rcos\theta$$ and $$v=rsin\theta$$ and $$dA=rdrd\theta$$, I have:
$$\int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \frac{r}{\sqrt{4+r^2}}drd\theta$$

And then solving the above will give me $$2(\sqrt{5}-2)\pi$$. 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
$$2(\sqrt{5}-2)\pi\leq\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA$$
when all I have now is:
$$2(\sqrt{5}-2)\pi=\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA$$

Last edited: Apr 9, 2009
2. Apr 10, 2009

### kof9595995

I'm very sorry I still don't know how to use Latex, so I'll just give you a hint, it is wrong just let u=sinx, v=siny simply, if you do so you must change the coordinate using Jacobian,
indeed, you just need a tiny correction, notice that x>sinx, y>siny, then replace sinx, siny by x and y, you get an inequality.
for pi/2,just find the upper bond of the integrand, it's quite straightforward.

3. Apr 10, 2009

### kof9595995

well, yeah, another correction x^2>sin^2(x),this would work.And besides upper bound we still need the lower bound ,right?
And mostly I was explaining this double integral does not equal to 2(sqrt5-2)pi,but larger than that

4. Apr 10, 2009

### gammamcc

uuuhhh, is that a typo? I just took your approach as correct. Yikes! Anyway x^2 \geq \sin^2x, etc. Sorry to previous poster....

5. Apr 11, 2009

### caseyjay

Hi,

I still don't quite get you. Basically you are letting x^2 > sin^2(x) and y^2 > sin^2(y). Why do you need to set it to greater and not let them be equal? And how can that lead to inequality? Is there some websites or book I can read about this?

6. Apr 11, 2009

### caseyjay

Hi,

So instead of letting x^2=sin^2(x), we can let x^2>sin^2(x)? can you explain to me in more details?

Thanks!

7. Apr 11, 2009

### kof9595995

well...it's not hard but a bit complicated to explain:
first, it is a definite integral an it has a certain value, but probably you cannot work it out, so it asks you for a range;
second, when you use u=sinx, v=siny, that is fine, but the following steps are wrong, you change the variables, so you need also change the integral element, for example, when you use polar coordinate, x=r*cosa, y=r*sina, you also change dxdy into rdrda,not just drda,but you just used dudv to replace dxdy at first, indeed this requires a Jacobian, you can search it on Wikipedia.
but using x^2>sin^2(x), you are not changing variables, but just construct an inequality in the original coordinate, so you won't need to change the integral element dxdy.
third, using x^2>sin^2(x) to construct the inequality, you'll find a new integrand always smaller than the original integrand, then you do the integral, you'll find the lower bond of the integral

8. Apr 11, 2009

### caseyjay

Thanks kof9595995.

How will using x^2>=sin^2(x), I will get a new integrand smaller than the original one? I'm lost on this.

Secondly, even if I manage to find the lower bound, how do I approach to find the upper bound?

Finally if there is any website or book that discussed about integral inequality, please let me know.

Thanks.

9. Apr 11, 2009

### kof9595995

sqrt[4+x^2+y^2]>sqrt[4+sin^2(x)+sin^2(y)], and it's the denominator ,so the integrand is smaller
the upper bound is rather easy, note that integrand is always smaller than 1/2, and the integrating area is pi*R^2=pi,so the integral must be smaller than pi/2

10. Apr 11, 2009

### caseyjay

Oh dear I think my calculus is really lousy. :( I'm so confused now.

How is the integrand be smaller than 1/2 and how will that lead to upper bound being pi/2?

I'm sorry I am really lost on this.

11. Apr 11, 2009

### kof9595995

because sqrt[4+sin^2(x)+sin^2(y)]>sqrt4=2, and if you think geometrically, the integral means a volume smaller than a cylinder with height 1/2 radius 1, algebraically, you can replace the integrand by 1/2, it'll also give the same result, but thinking geometrically is a shortcut.
Don't be upset,, it's still not abstract at this stage, just think more and practice more.

12. Apr 11, 2009

### caseyjay

Erm..... please let me know if I am on the right track.

the biggest value of sin^2(x) is 1. Hence, the biggest value of sqrt(4+sin^2(x)+sin^2(y)) is sqrt(6).

But why did you compare it with sqrt(4)?

Also I know that region A is defined by x^2+y^2=1, hence it's radius is 1. But why did you say that it has a height of 1/2 and not anything else?

13. Apr 11, 2009

### kof9595995

because you want one upper bond of the integrand, so you need 4+sin^2(x)+sin^2(y)>=4+0+0=4, so integrand<=1/sqrt4=1/2, that why I say the height is 1/2

14. Apr 12, 2009

### caseyjay

Hi kof9595995,

Thanks for your help. I more or less what is going on. Basically to find the upper bound, you let sin^2(x) and sin^2(y) be 0. But why do you not let sin^2(x) and sin^2(y) be 1 in order to find the lower bound?

Why do we proceed to convert the integrand to polar coordinates for lower bound, yet we let let sin^2(x) and sin^2(y) be 0 for upper bound?

15. Apr 12, 2009

### kof9595995

huh，it doesn't really matter but then you can't get the result, which the problem asks for

16. Apr 23, 2009

### caseyjay

Sorry for the late reply.

Thank you very much for your help!

Regards,
Casey