# Math Help - Compute Perimeter & Area

1. ## Compute Perimeter & Area

Consider the region in the plane enclosed by y = x^2 and y = 4.
a) Compute it's perimeter P and area A, and the ratio Q = A / P^2. (By squaring the perimeter we make this ratio independent of the unit length chosen to measure the region.)

b) Compare this ratio Q = A / P^2 among four different figures: the region in a), a square, a circle, and an equilateral triangle.

I found the area to be 16...... Not sure if that is correct. But I can't seem to figure out the perimeter. The rest seems to be comparing these #'s in different fashions, and using different shapes to compare your result to. Can someone help me out. TIA!!!

2. Originally Posted by TreeMoney
Consider the region in the plane enclosed by y = x^2 and y = 4.
a) Compute it's perimeter P and area A, and the ratio Q = A / P^2. (By squaring the perimeter we make this ratio independent of the unit length chosen to measure the region.)
The beautiful hand drawn picture below shows the dreadful situation.

The intersections points are,
$x^2=4$ thus, $x=\pm 2$.

Now we find the perimeter below and perimeter above.

1)Perimeter above is easy, simply $4$.

2)The Perimeter below is,
$\int_{-2}^2 \sqrt{1+[(x^2)']^2}dx$
Thus,
$\int_{-2}^2 \sqrt{1+(2x)^2}dx$
Let,
$u=2x$ then subsitution rule yields,
$\frac{1}{2}\int_{-4}^4 \sqrt{1+u^2} du$

Ignoring the limits of integration let us find the anti-derivative of,
$\int \sqrt{1+u^2} du$
Express as,
$\int \frac{(1+u^2)^2}{1} \cdot \frac{1}{1+u^2}du$
Let,
$t=\tan^{-1}(u)$ then, $t'=\frac{1}{1+u^2}$
Also,
$\sec^2 t=1+u^2$ then, $\sec^4 t=(1+u^2)^2$
Thus,
$\int \sec^4 t t' du=\int \sec^4 t dt$
Can you take it from here?
Integrate that then substitute $t=\tan^{-1}(u)$ once you finish.

3. Hello, TreeMoney!

Consider the region in the plane enclosed by $y = x^2$ and $y = 4$

a) Compute its perimeter $P$ and area $A$, and the ratio $Q = \frac{A}{P^2}$
Code:
                 4|
(-2,4)* - - - - + - - - - *(2,4)
|:::::::::|:::::::::|
|*::::::::|::::::::*|
| *:::::::|::::::*  |
|    *::::|::::*    |
--+--------***--------+--
-2         |         2

The area is: . $A \;= \;\int^2_{\text{-}2}\left(4 - x^2\right)\,dx \;=\;4x - \frac{x^3}{3}\,\bigg|^2_{\text{-}2} \;= \;\frac{32}{3}$

Since $y' = 2x$ and $1 + (y')^2 \:=\:1 + 4x^2$,
. . the arc length is: . $L \;=\;\int^2_{-2}\sqrt{1 + 4x^2}\,dx$

Let $2x = \tan\theta\quad\Rightarrow\quad dx = \frac{1}{2}\sec^2\theta\,d\theta$

. . The integral becomes: . $\int\sec^2\theta\left(\frac{1}{2}\sec\theta\,d\the ta\right) \;=\;\frac{1}{2}\int\sec^3\theta\,d\theta$

Using integration by parts or a standard formula, we have:
. . $\frac{1}{4}\left(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\right)$

Back-substitute: . $\frac{1}{4}\left(2x\sqrt{1+4x^2} + \ln\left|2x + \sqrt{1+4x^2}\right|\right)\,\bigg|^2_{\text{-}2}$

. . $= \;\frac{1}{4}\bigg[\left(4\sqrt{17} + \ln\left|4 + \sqrt{17}\right|\right) - \left(-4\sqrt{17} + \ln\left|-4 + \sqrt{17}\right|\right)\bigg]$

. . $= \;\frac{1}{4}\bigg[4\sqrt{17} + \ln\left|\sqrt{17} + 4\right| + 4\sqrt{17} - \ln\left|\sqrt{17} - 4\right|\bigg]$ $= \;\frac{1}{4}\left[8\sqrt{17} + \ln\left|\frac{\sqrt{17} + 4}{\sqrt{17} - 4}\right|\right]$

That fraction can be rationalized: . $\frac{\sqrt{17} + 4}{\sqrt{17} - 4}\cdot\frac{\sqrt{17} + 4}{\sqrt{17}+4} \:=\:\frac{(\sqrt{17} + 4)^2}{17 - 16}\;=\;(\sqrt{17} + 4)^2$

The answer becomes: . $\frac{1}{4}\left[8\sqrt{17} + \ln(\sqrt{17}+4)^2\right] \;= \;\frac{1}{4}\left[8\sqrt{17} + 2\ln(4 + \sqrt{17})\right]$

Hence, the arc length is: . $L \;=\;2\sqrt{17} + \frac{1}{2}\ln(4 + \sqrt{17})$

Therefore, the perimeter is: . $P \;= \;2\sqrt{17} + \frac{1}{2}\ln(4 + \sqrt{17}) + 4$

You take it from here . . .

4. Thanks guys!!! I don't know what I would do without everyone here!!!! Thanks again