My question is about the integration limits of a double integral

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I have to solve the following exercise :

Find the area of the $R$ region that is enclosed by the parable $y=x^2$ and the line $y=x+2$.

I did the following shape:

And I solved the double integral:

(the integration limits to $x$ axe: $y=x^2$ to $y=x+2$ and $-1leq xleq 2$ )

$A= int_R^int_dA=int_-1^2int_x^2^x+2dydx=int_-1^2[y]_x^2^x+2dx=int_-1^2(x+2-x^2)dx=[fracx^22]_-1^2+[2x]_-1^2-[fracx^33]_-1^2=frac92 $

If we reverse the sequence of integration (from $dydx$ to $dxdy$), I am confused about what the integration limits are

multiple-integral

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asked Jul 27 at 9:32

Deppie3910

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I have to solve the following exercise :

Find the area of the $R$ region that is enclosed by the parable $y=x^2$ and the line $y=x+2$.

I did the following shape:

And I solved the double integral:

(the integration limits to $x$ axe: $y=x^2$ to $y=x+2$ and $-1leq xleq 2$ )

$A= int_R^int_dA=int_-1^2int_x^2^x+2dydx=int_-1^2[y]_x^2^x+2dx=int_-1^2(x+2-x^2)dx=[fracx^22]_-1^2+[2x]_-1^2-[fracx^33]_-1^2=frac92 $

If we reverse the sequence of integration (from $dydx$ to $dxdy$), I am confused about what the integration limits are

multiple-integral

share|cite|improve this question

asked Jul 27 at 9:32

Deppie3910

205

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I have to solve the following exercise :

Find the area of the $R$ region that is enclosed by the parable $y=x^2$ and the line $y=x+2$.

I did the following shape:

And I solved the double integral:

(the integration limits to $x$ axe: $y=x^2$ to $y=x+2$ and $-1leq xleq 2$ )

$A= int_R^int_dA=int_-1^2int_x^2^x+2dydx=int_-1^2[y]_x^2^x+2dx=int_-1^2(x+2-x^2)dx=[fracx^22]_-1^2+[2x]_-1^2-[fracx^33]_-1^2=frac92 $

If we reverse the sequence of integration (from $dydx$ to $dxdy$), I am confused about what the integration limits are

multiple-integral

share|cite|improve this question

asked Jul 27 at 9:32

Deppie3910

205

I have to solve the following exercise :

Find the area of the $R$ region that is enclosed by the parable $y=x^2$ and the line $y=x+2$.

I did the following shape:

And I solved the double integral:

(the integration limits to $x$ axe: $y=x^2$ to $y=x+2$ and $-1leq xleq 2$ )

$A= int_R^int_dA=int_-1^2int_x^2^x+2dydx=int_-1^2[y]_x^2^x+2dx=int_-1^2(x+2-x^2)dx=[fracx^22]_-1^2+[2x]_-1^2-[fracx^33]_-1^2=frac92 $

If we reverse the sequence of integration (from $dydx$ to $dxdy$), I am confused about what the integration limits are

multiple-integral

share|cite|improve this question

asked Jul 27 at 9:32

Deppie3910

205

share|cite|improve this question

share|cite|improve this question

asked Jul 27 at 9:32

Deppie3910

205

asked Jul 27 at 9:32

Deppie3910

205

asked Jul 27 at 9:32

Deppie3910

205

205

add a comment | 

add a comment | 

3 Answers
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accepted

You have to split the region into two parts. For $y$ between $0$ and $1$ we have $-sqrt y <x <sqrt y$ and for $1<y<4$ it is $y-2<x<sqrt y$.

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answered Jul 27 at 9:37

Kavi Rama Murthy

20k2829

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The possible values for $y$ are the numbers from the interval $[0,4]$. If you reverse the order, then you get the sum of two integrals:$$int_0^1int_-sqrt y^sqrt y,mathrm dx,mathrm dy+int_1^4int_y-2^sqrt y,mathrm dx,mathrm dy.$$

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answered Jul 27 at 9:39

José Carlos Santos

113k1696173

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up vote
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If you reverse the order of integration you have two integrals to .$$ A=int_0^1int_-sqrt y^sqrt ydxdy+int_1^4int_y-2^sqrt ydxdy$$

You can take over from here and compare your result with your first attempt.

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answered Jul 27 at 9:45

Mohammad Riazi-Kermani

27.3k41851

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

You have to split the region into two parts. For $y$ between $0$ and $1$ we have $-sqrt y <x <sqrt y$ and for $1<y<4$ it is $y-2<x<sqrt y$.

share|cite|improve this answer

answered Jul 27 at 9:37

Kavi Rama Murthy

20k2829

add a comment | 

up vote
2
down vote



accepted

You have to split the region into two parts. For $y$ between $0$ and $1$ we have $-sqrt y <x <sqrt y$ and for $1<y<4$ it is $y-2<x<sqrt y$.

share|cite|improve this answer

answered Jul 27 at 9:37

Kavi Rama Murthy

20k2829

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

You have to split the region into two parts. For $y$ between $0$ and $1$ we have $-sqrt y <x <sqrt y$ and for $1<y<4$ it is $y-2<x<sqrt y$.

share|cite|improve this answer

answered Jul 27 at 9:37

Kavi Rama Murthy

20k2829

You have to split the region into two parts. For $y$ between $0$ and $1$ we have $-sqrt y <x <sqrt y$ and for $1<y<4$ it is $y-2<x<sqrt y$.

share|cite|improve this answer

answered Jul 27 at 9:37

Kavi Rama Murthy

20k2829

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 9:37

Kavi Rama Murthy

20k2829

answered Jul 27 at 9:37

Kavi Rama Murthy

20k2829

answered Jul 27 at 9:37

Kavi Rama Murthy

20k2829

20k2829

add a comment | 

add a comment | 

up vote
2
down vote

The possible values for $y$ are the numbers from the interval $[0,4]$. If you reverse the order, then you get the sum of two integrals:$$int_0^1int_-sqrt y^sqrt y,mathrm dx,mathrm dy+int_1^4int_y-2^sqrt y,mathrm dx,mathrm dy.$$

share|cite|improve this answer

answered Jul 27 at 9:39

José Carlos Santos

113k1696173

add a comment | 

up vote
2
down vote

The possible values for $y$ are the numbers from the interval $[0,4]$. If you reverse the order, then you get the sum of two integrals:$$int_0^1int_-sqrt y^sqrt y,mathrm dx,mathrm dy+int_1^4int_y-2^sqrt y,mathrm dx,mathrm dy.$$

share|cite|improve this answer

answered Jul 27 at 9:39

José Carlos Santos

113k1696173

add a comment | 

up vote
2
down vote

up vote
2
down vote

The possible values for $y$ are the numbers from the interval $[0,4]$. If you reverse the order, then you get the sum of two integrals:$$int_0^1int_-sqrt y^sqrt y,mathrm dx,mathrm dy+int_1^4int_y-2^sqrt y,mathrm dx,mathrm dy.$$

share|cite|improve this answer

answered Jul 27 at 9:39

José Carlos Santos

113k1696173

The possible values for $y$ are the numbers from the interval $[0,4]$. If you reverse the order, then you get the sum of two integrals:$$int_0^1int_-sqrt y^sqrt y,mathrm dx,mathrm dy+int_1^4int_y-2^sqrt y,mathrm dx,mathrm dy.$$

share|cite|improve this answer

answered Jul 27 at 9:39

José Carlos Santos

113k1696173

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 9:39

José Carlos Santos

113k1696173

answered Jul 27 at 9:39

José Carlos Santos

113k1696173

answered Jul 27 at 9:39

José Carlos Santos

113k1696173

113k1696173

add a comment | 

add a comment | 

up vote
2
down vote

If you reverse the order of integration you have two integrals to .$$ A=int_0^1int_-sqrt y^sqrt ydxdy+int_1^4int_y-2^sqrt ydxdy$$

You can take over from here and compare your result with your first attempt.

share|cite|improve this answer

answered Jul 27 at 9:45

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
2
down vote

If you reverse the order of integration you have two integrals to .$$ A=int_0^1int_-sqrt y^sqrt ydxdy+int_1^4int_y-2^sqrt ydxdy$$

You can take over from here and compare your result with your first attempt.

share|cite|improve this answer

answered Jul 27 at 9:45

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
2
down vote

up vote
2
down vote

If you reverse the order of integration you have two integrals to .$$ A=int_0^1int_-sqrt y^sqrt ydxdy+int_1^4int_y-2^sqrt ydxdy$$

You can take over from here and compare your result with your first attempt.

share|cite|improve this answer

answered Jul 27 at 9:45

Mohammad Riazi-Kermani

27.3k41851

If you reverse the order of integration you have two integrals to .$$ A=int_0^1int_-sqrt y^sqrt ydxdy+int_1^4int_y-2^sqrt ydxdy$$

You can take over from here and compare your result with your first attempt.

share|cite|improve this answer

answered Jul 27 at 9:45

Mohammad Riazi-Kermani

27.3k41851

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 9:45

Mohammad Riazi-Kermani

27.3k41851

answered Jul 27 at 9:45

Mohammad Riazi-Kermani

27.3k41851

answered Jul 27 at 9:45

Mohammad Riazi-Kermani

27.3k41851

27.3k41851

add a comment | 

add a comment | 

 

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