0
$\begingroup$

Evaluate the iterated integral $$\int_{-1}^1\int_0^{\sqrt{3+2y-y^2}}\cos\left(x^2+(y-1)^2\right)\,dy\,dx$$

Confused on how $y=0$ and $y=\sqrt{3+2y-y^2}$. Is this a typo or am I missing something?

$\endgroup$
  • $\begingroup$ Use $x=r\cos(\theta)$ and $y=1+r\sin(\theta)$ $\endgroup$ – DINEDINE Mar 16 '19 at 21:13
  • 2
    $\begingroup$ It looks to me that they switched the $dy$ and $dx$ when they typed the question. $\endgroup$ – user458276 Mar 16 '19 at 21:19
  • $\begingroup$ ok so it should be x = 0, x = sqrt(3+2y−y^2) ? $\endgroup$ – MasterYoshi Mar 16 '19 at 21:22
0
$\begingroup$

What are you confused about? As far as I can see the limits for the integral wrt $y,$ being $0$ and $\frac {1+\sqrt 7}{2}$ from the given conditions, are consistent with everything else.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Im a bit lost on how to find the theta and r. i cant seem to solve the equation by substituting in $rcostheta$ and $ rsintheta$ in $x^2=3+2y-y^2$ $\endgroup$ – MasterYoshi Mar 16 '19 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.