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:
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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 ......
The content structure of the provided text is a series of forum posts discussing a mathematical problem related to finding the area enclosed by two curves. The posts include the problem statement, multiple approaches to solving the problem, and a final solution.
multiple_questions_with_answers

1. the first post introduces the problem and asks for help.
2. the second post provides a solution to the problem, breaking it down into two integrals and calculating the area.
3. the third post offers an alternative approach to solving the problem.
4. the fourth post reiterates the problem and suggests breaking it into two integrals.
5. the fifth post confirms the approach and points out a minor error in the calculation.
6. the sixth post provides a detailed solution, including a graph and the final answer.
7. the seventh post clarifies the problem and provides another solution.
8. the eighth post confirms the solution and thanks the contributors.

each post contains a mathematical question or a step in solving the problem, and the final posts provide the answer.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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 ......
user
The content structure of the provided text is a series of forum posts discussing a mathematical problem related to finding the area enclosed by two curves. The posts include the problem statement, multiple approaches to solving the problem, and a final solution.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

1. The first post introduces the problem and asks for help.
2. The second post provides a solution to the problem, breaking it down into two integrals and calculating the area.
3. The third post offers an alternative approach to solving the problem.
4. The fourth post reiterates the problem and suggests breaking it into two integrals.
5. The fifth post confirms the approach and points out a minor error in the calculation.
6. The sixth post provides a detailed solution, including a graph and the final answer.
7. The seventh post clarifies the problem and provides another solution.
8. The eighth post confirms the solution and thanks the contributors.

Each post contains a mathematical question or a step in solving the problem, and the final posts provide the answer.