intergral by interpreting it in terms of areas

I am fairly new to this so...

$\displaystyle \int_{-3}^0(1+\sqrt{9- x^{2}})dx

\Rightarrow

\int_{-3}^{0}1dx + \int_{-3}^0\sqrt{9- x^{2}}dx

\Rightarrow

3 + \int_{-3}^0\sqrt{9- x^{2}}dx

$

assuming the form (from the reference in back of book)

$\displaystyle

\int_{}{}\sqrt{a^2- u^2}du

= \frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}{\frac{u}{a}+C

$

would be the way to solve the rest of it

but don't seem to get the answer of $\displaystyle 3+\frac{9\pi}{4}$

think I plugged in the right values finally

Quote:

Originally Posted by

**mr fantastic** Drawing a picture always helps. From your post title, I assume you're finding the area of the circle $\displaystyle (y - 1)^2 + x^2 = 3^2$ that lies in the third quadrant of the xy-plane ....

actually I don't know how you came up with that, but not doubting it

I think I plugged in the right values finally: (altho not sure about the x and C}

$\displaystyle

\frac{0}{2}\sqrt{-3^2-0} + \frac{-3^2}{2}\sin^{-1}{\left(\frac{0}{9}\right)} + C

\Rightarrow

0+\frac{9}{2}\frac{\pi}{2}

\Rightarrow

9\frac{\pi}{4}

$

so the final answer would be

$\displaystyle 3+9\frac{\pi}{4}$

btw I like the upgrades made to the MHF forum