# double integral

• Jul 23rd 2010, 02:23 PM
kientri123
double integral
My problem is like this:
Evaluate the double integral of x(1+y^2)^(-1/2)dA where R is the region in the first quadrant enclosed by y=x^2, y=4 and x =0.

I try to solve as type I, that is I set the limit of integration with respect to y from x^2 to 4 and then limit of ontegration with respect to x from 0 to 2.

I use trigonometric subtitution which yields

y=tan(delta) and dy=sec^2(delta)d(delta)

then I dont know how to change the limit integration which respect to y into the limit integration which respect to delta.

Thanks

P/s: this is my attached fomula:Attachment 18304
• Jul 23rd 2010, 02:36 PM
yeKciM
did u draw how it looks :D ?
and when u do that ... u'll see how u should do integration.... limit on x (dx) and y (dy) depending how u gonna do it... first by x or y ?

P.S. U don't need that subtitution to solve this one ...

if u change order of integration it's much much easier to solve .... :D look at it like this :D

$I= \int_{0} ^{4} dy \int_{0}^{\sqrt{y}} \frac{x}{\sqrt{1+y^2}}dx$

$I= \int_{0} ^{4} \frac{dy}{\sqrt{1+y^2}} \int_{0}^{\sqrt{y}}xdx$

$I=\frac {1}{2} \int_{0} ^{4} \frac {ydy}{\sqrt{1+y^2}}$

$1+y^2=t^2 \to 2ydy=2tdt \to ydy =tdt$

$I=\frac {1}{2} \int_{0} ^{4} \frac {dt}{t}$ ....
• Jul 23rd 2010, 06:50 PM
kientri123
Quote:

Originally Posted by yeKciM
did u draw how it looks :D ?
and when u do that ... u'll see how u should do integration.... limit on x (dx) and y (dy) depending how u gonna do it... first by x or y ?

P.S. U don't need that subtitution to solve this one ...

if u change order of integration it's much much easier to solve .... :D look at it like this :D

$I= \int_{0} ^{4} dy \int_{0}^{\sqrt{y}} \frac{x}{\sqrt{1+y^2}}dx$
$I= \int_{0} ^{4} \frac{dy}{\sqrt{1+y^2}} \int_{0}^{\sqrt{y}}xdx$
$I=\frac {1}{2} \int_{0} ^{4} \frac {ydy}{\sqrt{1+y^2}}$
$1+y^2=t^2 \to 2ydy=2tdt \to ydy =tdt$
$I=\frac {1}{2} \int_{0} ^{4} \frac {dt}{t}$ ....