# [SOLVED] stuck on area/integrals

• Sep 8th 2008, 07:33 PM
Legendsn3verdie
[SOLVED] stuck on area/integrals
http://i176.photobucket.com/albums/w...ntitled3-2.jpg
http://i176.photobucket.com/albums/w...ntitled6-1.jpg

can someone help me i know i m doing something wrong.. where did i start to go wrong?
• Sep 8th 2008, 09:25 PM
Shyam
Quote:

Originally Posted by Legendsn3verdie
http://i176.photobucket.com/albums/w...ntitled3-2.jpg
http://i176.photobucket.com/albums/w...ntitled6-1.jpg

can someone help me i know i m doing something wrong.. where did i start to go wrong?

Enclosed area $\displaystyle = \int\limits_0^1 {\sqrt x } ~dx + \int\limits_1^3 {\sqrt {\frac{{3 - x}}{2}} } ~dx \hfill \\$

Now, $\displaystyle \int\limits_0^1 {\sqrt x } ~dx = \left[ {\frac{2} {3}x^{\frac{3} {2}} } \right]_0^1 = \frac{2} {3} \hfill \\$

$\displaystyle {\text{Now, }}\int\limits_1^3 {\sqrt {\frac{{3 - x}} {2}} } dx \hfill \\$

$\displaystyle {\text{Let }}\frac{{3 - x}} {2} = u{\text{ , }}\frac{{ - dx}} {2} = du,{\text{ }} \Rightarrow dx = - 2du,{\text{ as }}x \to 1,{\text{ }}u \to 1{\text{ and as }}x \to 3,{\text{ }}u \to 0 \hfill \\$

$\displaystyle \int\limits_1^3 {\sqrt {\frac{{3 - x}} {2}} } ~dx = - 2\int\limits_1^0 {\sqrt u } ~du \hfill \\$

$\displaystyle = - 2\left[ {\frac{2} {3}u^{\frac{3} {2}} } \right]_1^0 = \frac{4} {3} \hfill \\$

$\displaystyle {\text{Enclosed area }} = \frac{2} {3} + \frac{4} {3} = \frac{6} {3} = 2 \hfill \\$
• Sep 8th 2008, 09:29 PM
mr fantastic
Quote:

Originally Posted by Legendsn3verdie
http://i176.photobucket.com/albums/w...ntitled3-2.jpg
http://i176.photobucket.com/albums/w...ntitled6-1.jpg

can someone help me i know i m doing something wrong.. where did i start to go wrong?

The enclosed area is exactly the same as the right hand side area between the inverses of each curve and the y-axis.

That is, the enclosed area is equal to the right hand side area between $\displaystyle y = x^2$, $\displaystyle y = 3 - 2x^2$ and the y-axis:

Enclosed area $\displaystyle = \int_{0}^{1} (3 - 2x^2) - x^2 \, dx$.

By the way. Don't bump.
• Sep 8th 2008, 09:29 PM
11rdc11
Make 2 integrals

$\displaystyle \int_0^1 \sqrt{x}~dx$ + $\displaystyle \int_1^3 \sqrt{\frac{3-x}{2}}~dx$
• Sep 8th 2008, 09:31 PM
Jhevon
Quote:

Originally Posted by Legendsn3verdie
I know this looks like horror but i really need help.

yes, you add the two integrals for the total area

(you messed up the second one at the end. the value of that integral is 4/3)

EDIT: geez! i'm really late. i'm tired anyway, i should go to bed...
• Sep 8th 2008, 09:31 PM
Soroban
Hello, Legendsn3verdie!

Quote:

Find the area enclosed by: .$\displaystyle x-y^2\:=\:0$ and $\displaystyle x + 2y^2 \:=\:3$
The graph looks like this:
Code:

              |         *    |            *               *      *               |  *               | *:::*               |*:::::*               |:::::::-       - - - - *:-:-:-:* - - - -               |:::::::               |*:::::*               | *:::*               |  *               *      *         *    |            *               |

It is easier integrate "sideways" ... with respect to $\displaystyle y.$

We have: .$\displaystyle x \:=\:y^2,\;\;x \:=\:3-y^2$

They intersect when: .$\displaystyle y^2 \:=\:3-2y^2 \quad\Rightarrow\quad 3y^2 \:=\:3 \quad\Rightarrow\quad y^2 \:=\:1$
. . They intersect at: $\displaystyle y \:=\:\pm1$

Formula: . $\displaystyle \text{Area} \;=\;\int^b_a\bigg[x_{\text{right}} - x_{\text{left}}\bigg]\,dy$

Due to the symmtry of the region, we can integrate from $\displaystyle y = 0$ to $\displaystyle y = 1$
. . and double the result.

We have: .$\displaystyle A \;\;=\;\;2\int^1_0\bigg[(3-2y^2) - y^2\bigg]\,dy \;=\;2\int^1_0\bigg[3 - 3y^2\bigg]\,dy \;\;=\; \;2\left(3y - y^3\right)\,\bigg]^1_0$

. . . . . $\displaystyle = \;\;2(3\cdot1-1^3) - 2(3\cdot0 - 0^3) \;\;=\;\;2(3-1) \;\;=\;\;\boxed{4}$

• Sep 8th 2008, 09:32 PM
Legendsn3verdie
Quote:

Originally Posted by Shyam
$\displaystyle \int\limits_1^3 {\sqrt x } dx = \left[ {\frac{2} {3}x^{\frac{3} {2}} } \right]_0^1 = \frac{2} {3} \hfill \\$

$\displaystyle {\text{Now, }}\int\limits_1^3 {\sqrt {\frac{{3 - x}} {2}} } dx \hfill \\$

$\displaystyle {\text{Let }}\frac{{3 - x}} {2} = u{\text{ , }}\frac{{ - dx}} {2} = du,{\text{ }} \Rightarrow dx = - 2du,{\text{ as }}x \to 1,{\text{ }}u \to 1{\text{ and as }}x \to 3,{\text{ }}u \to 0 \hfill \\$

$\displaystyle \int\limits_1^3 {\sqrt {\frac{{3 - x}} {2}} } ~dx = - 2\int\limits_1^0 {\sqrt u } ~du \hfill \\$

$\displaystyle = - 2\left[ {\frac{2} {3}u^{\frac{3} {2}} } \right]_1^0 = \frac{4} {3} \hfill \\$

$\displaystyle {\text{Required area }} = \frac{2} {3} + \frac{4} {3} = \frac{6} {3} = 2 \hfill \\$

hm wow guess i was close kinda, i just realized that graph flips over the x axis to cause its +- = 1 cause of the square roots in the begging. so i guess that total area is really 4, but anyways what u showed me help. thanks!
• Sep 8th 2008, 09:34 PM
Chris L T521
Quote:

Originally Posted by Legendsn3verdie
http://i176.photobucket.com/albums/w...ntitled3-2.jpg
http://i176.photobucket.com/albums/w...ntitled6-1.jpg

can someone help me i know i m doing something wrong.. where did i start to go wrong?

If you are to find the area between $\displaystyle x-y^2=0$ and $\displaystyle x+2y^2=3$, we see that we have two parabolas $\displaystyle x=y^2$, and $\displaystyle x=3-2y^2$.

To find where they intersect, set these equations equal to each other:

$\displaystyle y^2=3-2y^2\implies 3y^2=3\implies y^2=1\implies y=\pm1$

If graphed correctly, we can see that the area integral should be written in terms of y:

$\displaystyle A=\int_{-1}^1 \left(3-2y^2-y^2\right)\,dy=3\int_{-1}^1\left(1-y^2\right)\,dy=6\int_0^1\left(1-y^2\right)\,dy$

Does this make sense?

If not, try to see how I set up my integral from the graph of these two functions:

http://img.photobucket.com/albums/v4...tuff/graph.jpg

--Chris
• Sep 8th 2008, 09:42 PM
mr fantastic
Quote:

Originally Posted by Soroban
Hello, Legendsn3verdie!

The graph looks like this:
Code:

              |         *    |            *               *      *               |  *               | *:::*               |*:::::*               |:::::::-       - - - - *:-:-:-:* - - - -               |:::::::               |*:::::*               | *:::*               |  *               *      *         *    |            *               |
It is easier integrate "sideways" ... with respect to $\displaystyle y.$

We have: .$\displaystyle x \:=\:y^2,\;\;x \:=\:3-y^2$

They intersect when: .$\displaystyle y^2 \:=\:3-2y^2 \quad\Rightarrow\quad 3y^2 \:=\:3 \quad\Rightarrow\quad y^2 \:=\:1$
. . They intersect at: $\displaystyle y \:=\:\pm1$

Formula: . $\displaystyle \text{Area} \;=\;\int^b_a\bigg[x_{\text{right}} - x_{\text{left}}\bigg]\,dy$

Due to the symmtry of the region, we can integrate from $\displaystyle y = 0$ to $\displaystyle y = 1$
. . and double the result.

We have: .$\displaystyle A \;\;=\;\;2\int^1_0\bigg[(3-2y^2) - y^2\bigg]\,dy \;=\;2\int^1_0\bigg[3 - 3y^2\bigg]\,dy \;\;=\; \;2\left(3y - y^3\right)\,\bigg]^1_0$

. . . . . $\displaystyle = \;\;2(3\cdot1-1^3) - 2(3\cdot0 - 0^3) \;\;=\;\;2(3-1) \;\;=\;\;\boxed{4}$

My understanding is that only the upper half of the area is required ......