Double Integration

**brd_7** · February 22nd 2008, 02:23 PM

Evaluate the following double integrals over the given regions:

$\displaystyle\int\displaystyle\int_R \dfrac{x}{\sqrt{x^2 + y^2}}$ dA

$R= {(x,y) | 0 < x < \sqrt{4y -y^2}, 0 < y < 2}$

I have a feeling that this is not easily intergrated, im not too sure how i would go about doing it.. help would be appreciated.

Many thanks

**galactus** · February 22nd 2008, 02:32 PM

$\int\frac{x}{\sqrt{x^{2}+y^{2}}}dx=\sqrt{x^{2}+y^{ 2}}$

$|_{0}^{\sqrt{4y-y^{2}}}\frac{x}{\sqrt{x^{2}+y^{2}}}=2\sqrt{y}-y$

$\int[2\sqrt{y}-y]dy=\frac{4y^{\frac{3}{2}}}{3}-\frac{y^{2}}{2}$

Using the limits 0 to 2 we get $\frac{8\sqrt{2}}{3}-2$

**brd_7** · February 23rd 2008, 02:57 AM

I dont understand the first integration...

**mr fantastic** · February 23rd 2008, 03:24 AM

Originally Posted by brd_7

I dont understand the first integration...

To find $\int\frac{x}{\sqrt{x^{2}+y^{2}}} \, dx$ make the substitution $w = x^2 + y^2$ .

Remember that y is treated as a constant in this integral and so $\frac{dw}{dx} = 2x \Rightarrow dx = \frac{dw}{2x}$ .

Then the integral becomes $\int\frac{x}{\sqrt{w}} \, \frac{dw}{2x} = \frac{1}{2} \int\frac{1}{\sqrt{w}} \, dw = \frac{1}{2} \int w^{-1/2} \, dw = w^{1/2} = \sqrt{w}$ .

Now substitute back that $w = x^2 + y^2$ and you get $\sqrt{x^{2}+y^{2}}$ . Capisce?

**brd_7** · February 23rd 2008, 03:26 AM

Capisce

**Krizalid** · February 23rd 2008, 08:18 AM

Anyway, this is just chain rule.

You can modify the integrand such that you can find the hidden derivative, the rest follows.

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