# evaluate double intergral(s)

• Jan 5th 2011, 12:54 AM
Volga
evaluate double intergral(s)
I need a check on my solutions - I don't have answers in my study guide.

Also would appreciate commens on my 'train of thought' ie if there is a shorter way to arrive at my findings.

1. By changing the order of integration compute the integral:

$\int_0^1dx\int_x^{\sqrt{2x-x^2}}\frac{(x-1)e^y}{(y-1)}dy$

First, sketch the region of integration. Is there anyway to create sketches here?...For now, I'll just describe it:
- line through the origin y=x
- line x=1 parallel to Y axis
- curve $y=\sqrt{2x-x^2}$ which crosses and X axis (y=0) at x=0 and x=1, and crosses the line x=1 at y=1
- therefore the area of integration is defined by the line y=x at "the bottom" and the curve "at the top"

Second, change the order of variables. Some prelim work will be required to change the limits of the dx integral from the current limits of the dy integral. Need to find expression of x=f(y). For this, I will solve the quadratic equation and find its roots.

$y=\sqrt{2x-x^2}$
$y^2=2x-x^2$
$x^2-2x+y^2=0$
$D=4-4*1*y^2=4-4y^2$
$x=1\pm\sqrt{1-y^2}$

Difficulty. Which root (plus or minus) shall I pick for my purposes? both of them fit the original bound for the y function.

My solution. I will pick the root which 'fits' my plot, ie where if y=0 x=0, and this will be $1-\sqrt{1-y^2}$.

Next, the new limits of integration for dy integral are 0 and 1, from the observation of the plot.

Finally, the integration itself

$\int_0^1dx\int_x^{\sqrt{2x-x^2}}\frac{(x-1)e^y}{(y-1)}dy=\int_0^1dy\int_{1-\sqrt{1-y^2}}^y\frac{(x-1)e^y}{(y-1)}dx=$
$=\int_0^1dy \frac{e^y}{(y-1)}\int_{1-\sqrt{1-y^2}}^y(x-1)dx=$

$=\int_0^1(y^2-1)\frac{e^y}{(y-1)}dy=\int_0^1e^y(y+1)dy=(y+1)e^y\mid_0^1-\int_0^1e^ydy=e^yy\mid_0^1=e$

Note. I used partial integration for the dy integral. For dx integral, the workings are as follows:

$\int_{1-\sqrt{1-y^2}}^y(x-1)dx=\frac{(x-1)^2}{2}\mid_{1-\sqrt{1-y^2}}^y=\frac{(y-1)^2}{2}-\frac{(\sqrt{1-y^2})^2}{2}=y^2-1$
• Jan 5th 2011, 03:17 AM
HallsofIvy
Looks good to me! Congratulations!