Question

Evaluate the following integrals by converting to polar coordinates.
1 sqrt(1-x^2)
∫ ∫ e^(x^2+y^2)dydx
0 0

Please explain step by step

Answer 1

a) Conversion to polar coordinates: Let

I=∫11/2√∫x1−x2√1x2+y2−−−−−−√dydx.
If you draw a diagram of the situation you will see that

I=∫π/40∫r=secθr=11rrdrdθ=∫π/40secθ−1 dθ
=[12log∣∣∣1+sinθ1−sinθ∣∣∣−θ]π/40=log(1+2√)−π4.
(b) Standard hyperbolic substitution: To evaluate the inner integral set y=xsinhu and noting that sinh−1u=log(u+1+u2−−−−−√) we have

∫x1−x2√1x2+y2−−−−−−√dy=⎡⎣log⎛⎝yx+1+y2x2−−−−−−√⎞⎠⎤⎦x1−x2√
=log(1+2√)+logx−log(1+1−x2−−−−−√).(1)
Both the logs can be integrated by parts. The first is standard, ∫logx dx=xlogx−x+C and the second

∫log(1+1−x2−−−−−√) dx=xlog(1+1−x2−−−−−√)−x+sin−1x+C.
You will need (do the integration) to note that

11−x2−−−−−√−1=x2(1−x2)+1−x2−−−−−√.
And so upon integrating (1) between 1/2√ and 1 you obtain

I=(1−12√)log(1+2√)+[xlogx−xlog(1+1−x2−−−−−√)−sin−1x]11/2√
=log(1+2√)−π4.

Answer 2

its a hint

Answer 3

converting to polar coordinates is done with the formulas\[x=r\cos\theta\]\[y=r\sin\theta\]and in polar coordinates \[dA=rdrd\theta\]

Answer 4

the area suggested by the first bound is the regionwhich is a circle of radius \(r=1\)
therefor \(0\le r\le1\) and \(0\le\theta\le2\pi\)
the exponent changes thanks to our conversion as well:\[x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2(\sin^2\theta+\cos^2\theta)=r^2\]putting this all together we get our new integral for the area\[\large A=\int\int dA=\int_{0}^{2\pi}\int_{0}^{1}re^{r^2}drd\theta\]

Answer 5

actually I guess the last integral should be\[\large \int\int f(x)dA=\int_{0}^{2\pi}\int_{0}^{1}re^{r^2}drd\theta\]since you are integrating a function over the area

Answer 6

@Janex does this make sense?
can you do this integral?

Answer 7

it made sense but when i was integrating i get \[r (e-1) 2\pi\] not sure what i did wrong

Answer 8

\[\large\int_{0}^{2\pi}\int_{0}^{1}r e^{r^2}drd\theta=\int_{0}^{2\pi}\frac12e^{r^2}|_{0}^{1}d\theta=\frac12\int_{0}^{2\pi}e-1d\theta=\pi(e-1)\]do you see your mistake now?