text: 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 $\displaystyle = \int\limits_0^1 {\sqrt x } ~dx + \int\limits_1^3 {\sqrt {\frac{{3 - x}}{2}} } ~dx \hfill \\$
  
  Now, $\displaystyle \int\limits_0^1 {\sqrt x } ~dx = \left[ {\frac{2} {3}x^{\frac{3} {2}} } \right]_0^1 = \frac{2} {3} \hfill \\$
  
  $\displaystyle {\text{Now, }}\int\limits_1^3 {\sqrt {\frac{{3 - x}} {2}} } dx \hfill \\$
  
  $\displaystyle {\text{Let }}\frac{{3 - x}} {2} = u{\text{ , }}\frac{{ - dx}} {2} = du,{\text{ }} \Rightarrow dx = - 2du,{\text{ as }}x \to 1,{\text{ }}u \to 1{\text{ and as }}x \to 3,{\text{ }}u \to 0 \hfill \\$
  
  $\displaystyle \int\limits_1^3 {\sqrt {\frac{{3 - x}} {2}} } ~dx = - 2\int\limits_1^0 {\sqrt u } ~du \hfill \\$
  
  $\displaystyle = - 2\left[ {\frac{2} {3}u^{\frac{3} {2}} } \right]_1^0 = \frac{4} {3} \hfill \\$
  
  $\displaystyle {\text{Enclosed area }} = \frac{2} {3} + \frac{4} {3} = \frac{6} {3} = 2 \hfill \\$
  
  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 $\displaystyle y = x^2$, $\displaystyle y = 3 - 2x^2$ and the y-axis:
  
  Enclosed area $\displaystyle = \int_{0}^{1} (3 - 2x^2) - x^2 \, dx$.
  
  By the way. Don't bump.
  
  4. Make 2 integrals
  
  $\displaystyle \int_0^1 \sqrt{x}~dx$ + $\displaystyle \int_1^3 \sqrt{\frac{3-x}{2}}~dx$
  
  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: .$\displaystyle x-y^2\:=\:0$ and $\displaystyle x + 2y^2 \:=\:3$
  The graph looks like this:
  Code:
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  *     |             *
  *       *
  |   *
  | *:::*
  |*:::::*
  |:::::::-
  - - - - *:-:-:-:* - - - -
  |:::::::
  |*:::::*
  | *:::*
  |   *
  *       *
  *     |             *
  |
  
  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: .$\displaystyle A \;\;=\;\;2\int^1_0\bigg[(3-2y^2) - y^2\bigg]\,dy \;=\;2\int^1_0\bigg[3 - 3y^2\bigg]\,dy \;\;=\; \;2\left(3y - y^3\right)\,\bigg]^1_0$
  
  . . . . . $\displaystyle = \;\;2(3\cdot1-1^3) - 2(3\cdot0 - 0^3) \;\;=\;\;2(3-1) \;\;=\;\;\boxed{4}$
  
  7. 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!
  
  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 $\displaystyle x-y^2=0$ 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
  
  9. 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 ......
explanation: The provided text is a series of forum posts discussing a math problem related to finding the area enclosed by two curves. The posts contain a question and multiple answers from different users, each providing a step-by-step solution to the problem. The question is about finding the area enclosed by the curves \( x = y^2 \) and \( x = 3 - 2y^2 \).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 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 $\displaystyle = \int\limits_0^1 {\sqrt x } ~dx + \int\limits_1^3 {\sqrt {\frac{{3 - x}}{2}} } ~dx \hfill \\$
  
  Now, $\displaystyle \int\limits_0^1 {\sqrt x } ~dx = \left[ {\frac{2} {3}x^{\frac{3} {2}} } \right]_0^1 = \frac{2} {3} \hfill \\$
  
  $\displaystyle {\text{Now, }}\int\limits_1^3 {\sqrt {\frac{{3 - x}} {2}} } dx \hfill \\$
  
  $\displaystyle {\text{Let }}\frac{{3 - x}} {2} = u{\text{ , }}\frac{{ - dx}} {2} = du,{\text{ }} \Rightarrow dx = - 2du,{\text{ as }}x \to 1,{\text{ }}u \to 1{\text{ and as }}x \to 3,{\text{ }}u \to 0 \hfill \\$
  
  $\displaystyle \int\limits_1^3 {\sqrt {\frac{{3 - x}} {2}} } ~dx = - 2\int\limits_1^0 {\sqrt u } ~du \hfill \\$
  
  $\displaystyle = - 2\left[ {\frac{2} {3}u^{\frac{3} {2}} } \right]_1^0 = \frac{4} {3} \hfill \\$
  
  $\displaystyle {\text{Enclosed area }} = \frac{2} {3} + \frac{4} {3} = \frac{6} {3} = 2 \hfill \\$
  
  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 $\displaystyle y = x^2$, $\displaystyle y = 3 - 2x^2$ and the y-axis:
  
  Enclosed area $\displaystyle = \int_{0}^{1} (3 - 2x^2) - x^2 \, dx$.
  
  By the way. Don't bump.
  
  4. Make 2 integrals
  
  $\displaystyle \int_0^1 \sqrt{x}~dx$ + $\displaystyle \int_1^3 \sqrt{\frac{3-x}{2}}~dx$
  
  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: .$\displaystyle x-y^2\:=\:0$ 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: .$\displaystyle A \;\;=\;\;2\int^1_0\bigg[(3-2y^2) - y^2\bigg]\,dy \;=\;2\int^1_0\bigg[3 - 3y^2\bigg]\,dy \;\;=\; \;2\left(3y - y^3\right)\,\bigg]^1_0$
  
  . . . . . $\displaystyle = \;\;2(3\cdot1-1^3) - 2(3\cdot0 - 0^3) \;\;=\;\;2(3-1) \;\;=\;\;\boxed{4}$
  
  7. 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!
  
  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 $\displaystyle x-y^2=0$ 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
  
  9. 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 ......
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a series of forum posts discussing a math problem related to finding the area enclosed by two curves. The posts contain a question and multiple answers from different users, each providing a step-by-step solution to the problem. The question is about finding the area enclosed by the curves \( x = y^2 \) and \( x = 3 - 2y^2 \).
  
  Label: multiple-question-and-answer