# Boundaries and areas..not sure if I did it right.

• Feb 15th 2010, 05:56 PM
Jgirl689
Boundaries and areas..not sure if I did it right.
Consider the boundaries of the shaded region given (see attached graph).
http://www.webassign.net/cgi-bin/sym...ext%28-axis%29
http://www.webassign.net/cgi-bin/sym...pha%20%3D%2010
http://www.webassign.net/cgi-bin/sym...20root5%28x%29

Find the area A of this region by writing x as a function of y and integrating with respect to y.

I am not sure if I am doing this right, but this is what I did...

I took the derivative of 10 ^ 1/5 and assumed the integral was from 0 to 1 (since it shows on the graph I guess)
= 10^ 6/5 divided by 6/5 to get 50/6 x^ 1/5, plugged in the values and got 50/6, which is about 8.33 = area..is that right? But I'm confused to how to write it as a function. Is it like the integral (0,1) 50/6x ^ 6/5?
• Feb 15th 2010, 06:47 PM
Scott H
You are correct that the area under the curve is $\frac{50}{6}\approx 8.33$. The region in question is actually the region next to it, and the problem asks to integrate

\begin{aligned}
y&=10\sqrt[5]{x}\\
\sqrt[5]{x}&=\frac{y}{10}\\
x&=\frac{y^5}{10^5}
\end{aligned}

with respect to $y$. Note that the region extends from $y=0$ to $y=10$; therefore, our integral is

$\int_0^{10} \frac{y^5}{10^5}\,dy.$
• Feb 16th 2010, 01:16 PM
Jgirl689
^um I entered the answer in our homework section online and it says it's wrong :(. It has to be a fraction, or numerical value. I can't write that integral, but I substituted 0 and 10 to get a numerical value..wrong, then I tried to put in 0 and 1, still wrong! I don't know what I am doing wrong. Answer is not 8.33 either. Help!
• Feb 16th 2010, 01:28 PM
Scott H
Here is another hint:

$\int_0^{10}\frac{y^5}{10^5}\,dy=\frac{1}{10^5}\int _0^{10}y^5\,dy=\frac{1}{10^5}\left[\frac{y^6}{6}\right]_0^{10}.$
• Feb 16th 2010, 01:33 PM
Jgirl689
Quote:

Originally Posted by Scott H
Here is another hint:

$\int_0^{10}\frac{y^5}{10^5}\,dy=\frac{1}{10^5}\int _0^{10}y^5\,dy=\frac{1}{10^5}\left[\frac{y^6}{6}\right]_0^{10}.$

Ok so you then take the derivatie of it and then plug in the values? I am completly lost now.
• Feb 16th 2010, 01:49 PM
Jgirl689
or do you have to apply the chain rule too? Trying everything but still getting wrong answers..
• Feb 16th 2010, 01:53 PM
vince
is the answer $\frac5{3}$?...cuz that's what the following equals:
$
$
$\int_0^{10}\frac{y^5}{10^5}\,dy=\frac{1}{10^5}\int _0^{10}y^5\,dy=\frac{1}{10^5}\left[\frac{y^6}{6}\right]_0^{10} = \frac5{3}$

I haven't checked anything. i've just skimmed this thread and am trying to help you out by a)pointing out that your answer is the evaluation of the integral above (assuming the integral was set up right), and b)the actual computation of the integral leads to 5/3.
• Feb 16th 2010, 02:05 PM
Jgirl689
How did you get 5/3?? Aren't you supposed to plug in the integral values?
• Feb 16th 2010, 02:09 PM
icemanfan
That's what he did. The notation

$\frac{1}{10^5}\left[\frac{y^6}{6}\right]_0^{10}$

means

$\frac{1}{10^5}\left(\frac{10^6}{6} - \frac{0^6}{6}\right)$
• Feb 16th 2010, 02:16 PM
Jgirl689
I plugged that in and still got the wrong answer :(
• Feb 16th 2010, 02:30 PM
vince
Quote:

Originally Posted by Jgirl689
How did you get 5/3?? Aren't you supposed to plug in the integral values?

Ok silly...im gonna take this step by step.

The point of integration is to approximate, and in special cases, which happen far more often in textbooks and much less often in nature, the exact values of areas -- where "area" is used very broadly to mean the totality or total amount of something accumulated as a function "sweeps" through an interval of some kind.
In your case, it's very basic cause we're dealing with real numbers, and all we want is to compute the area bounded by two curves -- one of which isn't curved at all!

The only "twist" in your question is that it asks you to invert your description of the relation between the two variables of interest - one variables represents movement in the vertical dimension, the other movement in the horizontal direction. Surely, when trying to compute area, you want to move along one of the dimensions ("sweep"), if you're to accumulate the right amount.

Now the point of this problem is that it doesn't matter which direction you "sweep" through, and since when sweeping through the vertical dimension (as $y$ varies) allows you to deal with a simpler integrand (or the function that is getting "swept"), the problem explicitly asks you to do solve the integral that way. Isn't it easier to solve:

$\int_0^{10}\frac{y^5}{10^5}\,dy$

than this

$\int_0^{1}10-10x^{\frac1{5}}\,dx$

(well maybe not, they're both very easy once you learn ;))

They both will yield the same answer, but one is easier to work with. Again, to review, one integral is solved by "sweeping" in the vertical dimension $dy$ while the other in the horizontal $dx$-- they both will yield the same result.

So now all that's left is to convert the original function you're given
$y=f(x)=10x^{\frac1{5}}$, and convert it into one that is expressed as a function of $y$. By doing that, you're expressing the "sweep" in the right variables (the one in terms of $y$.)

Now go look at what Scott H has done above. He's laid out all the computational steps. I hope this "background" helps.
• Feb 16th 2010, 03:03 PM
Jgirl689
Ok so you have to solve the integral, problem is, we never taught how to do that :confused:
• Feb 16th 2010, 04:41 PM
vince
Quote:

Originally Posted by Jgirl689
Ok so you have to solve the integral, problem is, we never taught how to do that :confused:

we have solved the integral! the answer is $\frac5{3}$. if your answer book disagrees -- it's wrong. btw, i just edited one instance where i typed $x$ but meant $y$ (re-read if you wish :cool:)