# intergral by interpreting it in terms of areas

• Jun 28th 2010, 04:29 PM
bigwave
intergral by interpreting it in terms of areas
I am fairly new to this so...

$\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)
$
\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 $3+\frac{9\pi}{4}$
• Jun 28th 2010, 04:34 PM
mr fantastic
Quote:

Originally Posted by bigwave
I am fairly new to this so...

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

I do. So clearly there's a mistake in your algebra. If you post all your working your mistake can be pointed out.
• Jun 28th 2010, 04:45 PM
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.
• Jun 28th 2010, 04:57 PM
mr fantastic
Quote:

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 $(y - 1)^2 + x^2 = 3^2$ that lies in the second quadrant of the xy-plane ....
• Jun 28th 2010, 07:49 PM
bigwave
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 $(y - 1)^2 + x^2 = 3^2$ 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}

$
\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

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

• Jun 28th 2010, 08:56 PM
mr fantastic
Quote:

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

[snip]

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

Quote:

Originally Posted by bigwave
[snip]
I think I plugged in the right values finally: (altho not sure about the x and C}

$
\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

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

Sorry, but this working does not make a lot of sense to me. I expected to see an $\sin^{-1} (-1)$ 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.
• Jun 29th 2010, 07:07 AM
HallsofIvy
mr fantastic is answering the question you asked but it is the wrong question! You appear to have completely misunderstood this problem.

You titled this thread "Integral by interpreting it in terms of area" and you have made no attempt to that. The point of this problem is NOT to do any integration at all!

If $y= 1+ \sqrt{9- x^2}$, then $y- 1= \sqrt{9- x^2}$, $(y- 1)^2= 9- x^2$ or $x^2+ (y- 1)^2= 9$. That's a circle with center at (0, 1) and radius 3. Because the original equation used only the positive square root, $y= 1+ \sqrt{9- x^2}> 1$ 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.

What is the area of a circle of radius 3? What is the area of a quarter of that circle? What is the area of a rectangle of length 3 and height 1? What is the sum of those two areas?
• Jun 29th 2010, 11:19 AM
bigwave
thanks everyone for your help... just new to all this...