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i.e.\[V=2 \pi \int\limits_{0}^{1}r \sqrt{2-r^2}dr-2 \pi \int\limits_{0}^{1}r^3 dr\]

That is\[V_L=-\frac{2\pi}{3}(2-r^2)^{3/2}|_0^1=\frac{2 \pi}{3}(2\sqrt{2}-1)\]

The second component is\[V_R=-2\pi \frac{r^4}{4}|_0^1=-\frac{\pi}{2}\]

So\[V=V_L+V_R=\frac{2\pi}{3}(2\sqrt{2}-1)-\frac{\pi}{2}\]

thanks.u help me out.thanks my hero =)

thanks.u help me out.thanks my hero =)

thanks.u help me out.thanks my hero =)

and you're welcome :)

your answer is correct.but how come u get x^2 + y^2 =1?

yeah.i get it now.thanks lokisan =)

np :)