1. integral need help !

the integral (from 0 to a) of sqrt(a^2 -r^2) 2r dr.

Thank you very much.

2. $
\int_0^a \ {\sqrt{a^2-r^2} 2r dr} = \frac{2}{3}a^3
$

I might be wrong though

3. Good work. That's correct.

4. hi nertil1 ,

Thanks. Could you please show how you come up with this answer? Thank you very much.

5. Originally Posted by kittycat
hi nertil1 ,

Thanks. Could you please show how you come up with this answer? Thank you very much.
Use the substitution $t=a^2-r^2$

6. Let $r = a\sin \theta$.

7. $2\int_{0}^{a}r\sqrt{a^{2}-r^{2}}dr$

Let $r=asin{\theta}, \;\ dr=acos{\theta}d{\theta}$

$2\int_{0}^{a}{asin{\theta}\sqrt{a^{2}(1-sin^{2}{\theta})}acos{\theta}}d{\theta}$

Remember, $1-sin^{2}{\theta}=cos^{2}{\theta}$

$2a^{3}\int_{0}^{a}{sin{\theta}cos^{2}{\theta}}d{\t heta}$

$-2a^{3}\frac{cos^{3}{\theta}}{3}$

But, $\theta=sin^{-1}(\frac{r}{a})$

Sub back in and get:

$\frac{2(r^{2}-a^{2})\sqrt{a^{2}-r^{2}}}{3}$

Sub in the limits of integration and get:

$0-\frac{2a^{3}}{3}=\frac{2a^{3}}{3}$

8. Why use such a scary substitution? The derivative already appears on the outside.