0
$\begingroup$

Problem 1

Convert to polar form and solve

$$\int^{1}_{0}\int_{0}^{\sqrt{2y-y^2}}(1-x^2-y^2)\text{ dx dy}$$

$x^2+y^2-2y=0$

$x^2+(y-1)^2=1$

$x=rcos\theta$ $y=rsin\theta+1$

$r^2=1, r=1$

$$\int^{\pi}_{0}\int^{1}_{0}(1-r^2cos^2\theta-(y^2cos^2\theta+1)) rdrd\theta$$ $$\int^{\pi}_{0}\int^{1}_{0}(-r^3)drd\theta = -1/4{\pi}$$

$\endgroup$
2
  • 1
    $\begingroup$ What is the question? $\endgroup$
    – MSDG
    Mar 17, 2019 at 20:52
  • $\begingroup$ is this correct? i do not have the answer to this problem $\endgroup$ Mar 17, 2019 at 20:54

1 Answer 1

0
$\begingroup$

You made an error when you substituted $y^2$. For the reference, the correct result is $\frac{2}{3} - \frac{\pi}{8}$.

$\endgroup$
4
  • $\begingroup$ Hey, i dont know why i cannot see that mistake? you mean when i plugged $ y= rsin\theta+1$ in $y^2$? $\endgroup$ Mar 17, 2019 at 21:13
  • $\begingroup$ $y^2 = (r\sin\theta+1)^2$ and not what you did in the OP (if I'm reading this correctly). $\endgroup$
    – Klaus
    Mar 17, 2019 at 21:19
  • $\begingroup$ ohh, so $(1-r^2cos\theta-(r^2sin^2\theta+2rsin\theta+1))$? i get the integral of $-r^3-2r^2sin\theta$ ? $\endgroup$ Mar 17, 2019 at 21:44
  • $\begingroup$ the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect? $\endgroup$ Mar 17, 2019 at 22:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .