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 \\
$
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.
Hello, Legendsn3verdie!
The graph looks like this:Find the area enclosed by: .$\displaystyle x-y^2\:=\:0$ and $\displaystyle x + 2y^2 \:=\:3$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 $
. . . . . $\displaystyle = \;\;2(3\cdot1-1^3) - 2(3\cdot0 - 0^3) \;\;=\;\;2(3-1) \;\;=\;\;\boxed{4}$
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