anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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.
anonymous
  • anonymous
its a hint
TuringTest
  • TuringTest
converting to polar coordinates is done with the formulas\[x=r\cos\theta\]\[y=r\sin\theta\]and in polar coordinates \[dA=rdrd\theta\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TuringTest
  • TuringTest
the area suggested by the first bound is the region|dw:1333654243822:dw|which 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\[\large A=\int\int dA=\int_{0}^{2\pi}\int_{0}^{1}re^{r^2}drd\theta\]
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
@Janex does this make sense? can you do this integral?
anonymous
  • anonymous
it made sense but when i was integrating i get \[r (e-1) 2\pi\] not sure what i did wrong
TuringTest
  • TuringTest
\[\large\int_{0}^{2\pi}\int_{0}^{1}r e^{r^2}drd\theta=\int_{0}^{2\pi}\frac12e^{r^2}|_{0}^{1}d\theta=\frac12\int_{0}^{2\pi}e-1d\theta=\pi(e-1)\]do you see your mistake now?

Looking for something else?

Not the answer you are looking for? Search for more explanations.