Question

change the cartesian integral into an equivalent polar integral:

Answer 1

\[\int\limits_{\sqrt{2}}^{2}\int\limits_{\sqrt{4-y^{2}}}^{y} dx dy\]

Answer 2

i think this is what it looks like at least

Answer 3

remember that x=rcos(t)
y=rsin(t)
and r=sqrt(x^2+y^2), but thats really already given as 2

Answer 4

I think I understand how to get pi/4 < theta < pi/2

But finding the bounds for r is throwing me off.

Answer 5

well, since r only really changes from 0 to 2; the bounds for r are 0 to 2

Answer 6

it r changed according to some function of t

the r would be bound from 0 to f(t)

Answer 7

according to the answers, r is bounded by 2 < r < 2csc(theta)

Answer 8

but at pi/2 r can be 2 and rad(2)

Answer 9

it might help to see the limits in this fashion to be able to see how to change them out

\[\Large \int_{y=\sqrt{2}}^{y=2}~~~\int_{x=\sqrt{4-y^{2}}}^{x=y} ~~~dx~dy\]

Answer 10

my book says to plug in x=rcos(theta) etc and magically gets the polar bounds. They dont show any intermediate steps so im really confused.

Answer 11

\[x= \sqrt{r^2-y^2}\]\[2cos(t) = \sqrt{2^2-(2sin(t))^2}\]\[4cos^2(t) = 4-4sin^2(t)\]\[cos^2(t) +sin^2(t)=1~\text{for any t}\]

\[x= y\]\[2cos(t) = 2sin(t)\]\[t=\frac{pi}4\]

something along those lines i believe

Answer 12

y = 2
2sin(t) = 2
sin(t) = 1; when t=pi/2

y=sqrt(2)
2sin(t)=sqrt(2)
sin(t)=sqrt(2)/2; when t=pi/4