# A conversion of double integral to polar and evaluate, check

## Problem 1

Convert to polar form and solve

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

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

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

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

$$r^2=1, r=1$$

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

• What is the question? – MSDG Mar 17 '19 at 20:52
• is this correct? i do not have the answer to this problem – MasterYoshi Mar 17 '19 at 20:54

You made an error when you substituted $$y^2$$. For the reference, the correct result is $$\frac{2}{3} - \frac{\pi}{8}$$.
• Hey, i dont know why i cannot see that mistake? you mean when i plugged $y= rsin\theta+1$ in $y^2$? – MasterYoshi Mar 17 '19 at 21:13
• $y^2 = (r\sin\theta+1)^2$ and not what you did in the OP (if I'm reading this correctly). – Klaus Mar 17 '19 at 21:19
• ohh, so $(1-r^2cos\theta-(r^2sin^2\theta+2rsin\theta+1))$? i get the integral of $-r^3-2r^2sin\theta$ ? – MasterYoshi Mar 17 '19 at 21:44
• the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect? – MasterYoshi Mar 17 '19 at 22:16