double integrals

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March 1st 2006, 09:18 AM
vish1987

double integrals

an some1 please help me with the following 2 question, i have attemped both but seem to come out with the correct answer:

1. ∫dx ∫(x-y)dy the 1st integral sign has the limits a and 0. the 2nd sign has the limits y1 and 0. where y1 = (a^2 -x^2)^0.5
for this ques the answer should be 0.

2. by changing the order of integration evaluate:
∫dx∫y^-1 sinycos(x/y)dy for the 1st integral sign, limits are 1 and 0. the 2nd sign has limits 1 and x.
answer: sin1(1-cos1)
March 1st 2006, 09:59 AM
topsquark

Quote:

Originally Posted by vish1987

an some1 please help me with the following 2 question, i have attemped both but seem to come out with the correct answer:

1. ∫dx ∫(x-y)dy the 1st integral sign has the limits a and 0. the 2nd sign has the limits y1 and 0. where y1 = (a^2 -x^2)^0.5
for this ques the answer should be 0.

There may be a more elegant way, but I always do these in baby steps.

$\int_0^a dx \int_0^{\sqrt{a^2-x^2}} dy(x-y)$
$=\int_0^a dx(xy-(1/2)y^2)|_0^{\sqrt{a^2-x^2}}$
$=\int_0^a dx \, x \sqrt{a^2-x^2} \, - \, \int_0^a \, dx(1/2)(a^2-x^2)$

Now, the first integral:
Use $u=a^2-x^2$
$du=-2xdx$
$\int dx \, x \sqrt{a^2-x^2}=\int du(-1/2) \sqrt{u}=(-1/2)(2/3)u^{3/2}$
$=(-1/3)(a^2-x^2)^{3/2}$
Inserting the end limits:
$\int_0^a dx \, x \sqrt{a^2-x^2}=(-1/3)(a^2-x^2)|_0^a=(1/3)a^3$

The second integral is easy:
$\int_0^a \, dx(1/2)(a^2-x^2)=(1/2)[a^2x-(1/3)x^3]|_0^a$
$=(1/2)a^3-(1/6)a^3=(1/3)a^3$

So we need to subtract. Finally:
$\int_0^a dx \int_0^{\sqrt{a^2-x^2}} dy(x-y) =(1/3)a^3-(1/3)a^3=0$ .

-Dan
March 1st 2006, 10:16 AM
topsquark

Someone else is going to have to do 2). I can't figure it.

-Dan
March 1st 2006, 10:47 AM
vish1987

ok thanks for the 1st one neway!
March 1st 2006, 12:14 PM
CaptainBlack

Quote:

Originally Posted by vish1987

2. by changing the order of integration evaluate:
∫dx∫y^-1 sinycos(x/y)dy for the 1st integral sign, limits are 1 and 0. the 2nd sign has limits 1 and x.
answer: sin1(1-cos1)

The integral to be evaluated is:

$ \int_0^1 \int_x^1 \frac{\sin(y) \cos(x/y)}{y} dy\ dx $

Which we may write:

$ \int_A \frac{\sin(y) \cos(x/y)}{y} ds $

where A is the region between the y-axis, and the lines x=1 and x=y,
and ds the the area element. This now may be written as:

$ \int_0^1 \int_0^y \frac{\sin(y)\cos(x/y)}{y} dx\ dy $ ,

which may be integrated to give:

$ \int_0^1 \frac{\sin(y)}{y} \ [ y \sin(x/y)]_{x=0}^{x=y}}\ dy=$ $ \int_0^1 \sin(y) \sin(1) dy=\sin(1)[-\cos(y]_0^1=\sin(1)[1-\cos(1)] $

RonL