# Math Help - [SOLVED] stuck on area/integrals

1. ## [SOLVED] stuck on area/integrals

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

2. Originally Posted by Legendsn3verdie

can someone help me i know i m doing something wrong.. where did i start to go wrong?
Enclosed area $= \int\limits_0^1 {\sqrt x } ~dx + \int\limits_1^3 {\sqrt {\frac{{3 - x}}{2}} } ~dx \hfill \\$

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

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

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

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

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

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

3. Originally Posted by Legendsn3verdie

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 $y = x^2$, $y = 3 - 2x^2$ and the y-axis:

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

By the way. Don't bump.

4. Make 2 integrals

$\int_0^1 \sqrt{x}~dx$ + $\int_1^3 \sqrt{\frac{3-x}{2}}~dx$

5. 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...

6. Hello, Legendsn3verdie!

Find the area enclosed by: . $x-y^2\:=\:0$ and $x + 2y^2 \:=\:3$
The graph looks like this:
Code:
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|*:::::*
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- - - - *:-:-:-:* - - - -
|:::::::
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It is easier integrate "sideways" ... with respect to $y.$

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

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

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

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

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

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

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

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

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

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

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

${\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!

8. Originally Posted by Legendsn3verdie

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 $x-y^2=0$ and $x+2y^2=3$, we see that we have two parabolas $x=y^2$, and $x=3-2y^2$.

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

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

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

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

Does this make sense?

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

--Chris

9. Originally Posted by Soroban
Hello, Legendsn3verdie!

The graph looks like this:
Code:
              |
*     |             *
*       *
|   *
| *:::*
|*:::::*
|:::::::-
- - - - *:-:-:-:* - - - -
|:::::::
|*:::::*
| *:::*
|   *
*       *
*     |             *
|
It is easier integrate "sideways" ... with respect to $y.$

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

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

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

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

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

. . . . . $= \;\;2(3\cdot1-1^3) - 2(3\cdot0 - 0^3) \;\;=\;\;2(3-1) \;\;=\;\;\boxed{4}$
My understanding is that only the upper half of the area is required ......