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Thread: [SOLVED] stuck on area/integrals

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




    can someone help me i know i m doing something wrong.. where did i start to go wrong?
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    Quote Originally Posted by Legendsn3verdie View Post



    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 \\
    $
    Last edited by Shyam; Sep 9th 2008 at 11:14 AM.
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  3. #3
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    Quote Originally Posted by Legendsn3verdie View Post



    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.
    Last edited by mr fantastic; Sep 8th 2008 at 09:33 PM. Reason: Added the last line.
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  4. #4
    Super Member 11rdc11's Avatar
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    Make 2 integrals

    $\displaystyle \int_0^1 \sqrt{x}~dx$ + $\displaystyle \int_1^3 \sqrt{\frac{3-x}{2}}~dx$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    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...
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    Hello, Legendsn3verdie!

    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 $

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

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

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

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

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

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

    $\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!
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post



    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:



    --Chris
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  9. #9
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    Quote Originally Posted by Soroban View Post
    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 ......
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