# Thread: intergral by interpreting it in terms of areas

1. ## 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 $\displaystyle 3+\frac{9\pi}{4}$

2. Originally Posted by bigwave
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 $\displaystyle 3+\frac{9\pi}{4}$
I do. So clearly there's a mistake in your algebra. If you post all your working your mistake can be pointed out.

3. so I am using the right form just have to carefull what I do with it... i will repost after i go thru this again.

4. Originally Posted by bigwave
so I am using the right form just have to carefull what I do with it... i will repost after i go thru this again.
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 second quadrant of the xy-plane ....

5. ## think I plugged in the right values finally

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

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

6. Originally Posted by bigwave
actually I don't know how you came up with that, but not doubting it

[snip]
$\displaystyle y = 1 + \sqrt{9 - x^2} \Rightarrow y - 1 = \sqrt{9 - x^2} \Rightarrow (y - 1)^2 = 9 - x^2$ etc.

Originally Posted by bigwave
[snip]
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

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

Sorry, but this working does not make a lot of sense to me. I expected to see an $\displaystyle \sin^{-1} (-1)$ somewhere .... I also suggest you put brackets around negative numbers when squaring them (-3^2 is likely to be confused with -9 ....) Note also that the arbitrary constant is not required when finding definite integrals.
If $\displaystyle y= 1+ \sqrt{9- x^2}$, then $\displaystyle y- 1= \sqrt{9- x^2}$, $\displaystyle (y- 1)^2= 9- x^2$ or $\displaystyle x^2+ (y- 1)^2= 9$. That's a circle with center at (0, 1) and radius 3. Because the original equation used only the positive square root, $\displaystyle y= 1+ \sqrt{9- x^2}> 1$ so we really have the upper half disk. Finally, because the integral is from x=-3 to 0, while the entire circle goes from x=-3 to x= +3, this integral is really the area of a quarter circle plus the rectangular section between that and the x-axis.