Question

Integrate polynomials

Answer 1

??

Answer 2

i typed out then lost it\

Answer 3

i'll start again, sorry

Answer 4

\[\int\limits_{-1}^{1}\int\limits_{-\sqrt{1-x ^{2}}}^{\sqrt{1-x ^{2}}}\frac{ 2dydx }{ (1+x ^{2}+y ^{2})^{2} }\]

Answer 5

Evaluate the integrals by changing to polar coordinates.

Answer 6

i know r^2=y^2+x^2=1

Answer 7

so it's a cirlce radius 1

Answer 8

ok, you just need to substitute dy and dx in the integral in terms of polar coordinates. use y=r sin and x = r cos to differentiate...

Answer 9

i'm getting lost on the limits how do we determine those?

Answer 10

put x = rcos^2 theta. since r=1 you get x=cos^2theta

Answer 11

and it reduces to sin theta for upper limit and -sin theta for lower limit.

Answer 12

i know x=r cos theta and y=r sin theta

why are you saying x = rcos^2 theta

Answer 13

I meant x^2=cos^2 theta

Answer 14

how about conversion of the outer limits to polar angles?

Answer 15

the answer thay have given is pi

Answer 16

try

\[

\int \limits_0^{2\pi} \int \limits_0^1 \dfrac{2}{(1+r^2)^2} r dr d\theta
\]

Answer 17

you're integrating over a unit disk
so the bounds will be :
theta : 0->2pi
r : 0->1

Answer 18

how do we get that information from the question?

Answer 19

first, sketch the region represented by given bounds

Answer 20

\[\int\limits_{-1}^{1}\int\limits_{-\sqrt{1-x ^{2}}}^{\sqrt{1-x ^{2}}}\frac{ 2dydx }{ (1+x ^{2}+y ^{2})^{2} } \]

Answer 21

sketch below curves :
x= -1
x = 1
y = -sqrt(1-x^2)
y = sqrt(1+x^2)

Answer 22

the bounded region of above curves is the region over which you're integrating the integrand