# How can I evaluate this double integral?

The integral that I'm trying to evaluate is:

$$\int\limits_{x=0}^{2} \int\limits_{y=0}^{\frac{x^2}{2}} \frac{x}{(x^2+y^2+1)^{\frac{1}{2}}} dydx$$

I can get as far as

$$\int\limits_{x=0}^{2} x \int\limits_{y=0}^{\frac{x^2}{2}} \frac{1}{(x^2+y^2+1)^{\frac{1}{2}}} dydx$$

But I have no idea how to evaluate this integral.

Subbing in $c=(1+x^2)$ and focusing, I am working on:

$$\int \frac{1}{(c+y^2)^{\frac{1}{2}}} dy$$

but I can see no obvious way to proceed.

• Have you tried changing the coordinates? – 5xum Jan 27 '14 at 8:59
• Try $r^2 = x^2 + y^2$ – IAmNoOne Jan 27 '14 at 9:00

Hint: Draw a picture. Use it to change the order of integration, integrating first with respect to $x$, where $x$ travels from $\sqrt{2y}$ to $2$. The first integration will be an easy substitution.

• Why from $\sqrt{2y}$ to $2$ and then what would the limits on y be? I can't really draw in 3D – NictraSavios Jan 27 '14 at 9:19
• It is 2D. Draw the parabola $y=x^2/2$. Our given integral is over the region in the first quadrant, below the parabola, from $x=0$ to $x=2$. If we integrate with respect to $x$ first, then $x$ starts at the parabola, that is, at $x=\sqrt{2y}$, and goes to $2$. Then $y$ goes from $0$ to $2$. – André Nicolas Jan 27 '14 at 9:23
• Whoa - So its just a square? $x \in [0,2]$ and $y \in [0,2]$ ? – NictraSavios Jan 27 '14 at 9:27
• No, it is the region below $y=x^2/2$, above the $x$-axis, $x=0$ to $x=2$. So it is kind of triangular-looking, but with one edge curvy. – André Nicolas Jan 27 '14 at 9:29
• Okay, I see now. One more question, what does the integral mean? How does the function $\frac{x}{(1+x^2+y^2)^{(1/2)}}$ get related to $x^2/2$ ? – NictraSavios Jan 27 '14 at 9:33

You reveived very good answers from André Nicolas and Semsem and this is really what you have to do.

So, my contribution will be very modest : here is the result of the antiderivative which makes problem to you

$$\int \frac{1}{(c+y^2)^{\frac{1}{2}}} dy = \log \left(\sqrt{\text{c}+y^2}+y\right)$$

This is obtained changing variable $y = \sqrt{\text{c}} \sinh (z)$.

But, if you use this for continuing working your problem, I am afraid you will probably face nightmares.