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Calculating the volume with double integral

Finding volume of a shape using double integralFinding volumes - when to use double integrals and triple integrals?Double Integral Set Updouble integral to find the volumeDouble and Triple IntergralDouble Integral Application with DisksSolid volume using double integralQuestion about calculating the volume of a sphere using spherical coordinatesVolume with double integralintegration of double integral with euler number include it.

2

$begingroup$

Hello i am trying to calculate the volume for a double integral but i am having problem with define the integral because it is not given in a pure form. I have $z = xy$, $x+y+z=1$ $z=0$ my approach is to set the function for a integral to be $$int_Dxy$$ and to find the $limits for$ $dy$ i set $z$ to be zero it is also given by definition and i get $y = 1-x$ after that i set both $z$ and $y$ to zero and i get $x = 1$ so i have the followinglimits $$int_0^1 int_0^x-1xy$$ but i am not getting the right answer after evaluating the integral. What confusing me here is that the integral is not given by default here also the other thing that confuses me is i have the same problem but to be solved with triple integral. I am thinking maybe for the volume i just need $dydx$ without a function but i am not sure. Thank you for any help in advance.

calculus integration multivariable-calculus volume

share|cite|improve this question

edited May 27 at 7:26

Boris Borovski

asked May 27 at 7:24

Boris BorovskiBoris Borovski

556

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ...and the right answer is...?
  $endgroup$
  – DonAntonio
  May 27 at 7:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonAntonio I mean the approach i take its not right i get the sarcasm : )
  $endgroup$
  – Boris Borovski
  May 27 at 7:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bo Ok, fine...and then how do you know you're not getting "the right answer"??
  $endgroup$
  – DonAntonio
  May 27 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonAntonio When i evaluate the integral the and i substitute the limits the terms are cancelling out to 0 and the answer is $frac1712-2ln2$. I didn't solve a problem which i need to find just the volume.
  $endgroup$
  – Boris Borovski
  May 27 at 7:44
  

  
  
  
  

add a comment | 

2

$begingroup$

Hello i am trying to calculate the volume for a double integral but i am having problem with define the integral because it is not given in a pure form. I have $z = xy$, $x+y+z=1$ $z=0$ my approach is to set the function for a integral to be $$int_Dxy$$ and to find the $limits for$ $dy$ i set $z$ to be zero it is also given by definition and i get $y = 1-x$ after that i set both $z$ and $y$ to zero and i get $x = 1$ so i have the followinglimits $$int_0^1 int_0^x-1xy$$ but i am not getting the right answer after evaluating the integral. What confusing me here is that the integral is not given by default here also the other thing that confuses me is i have the same problem but to be solved with triple integral. I am thinking maybe for the volume i just need $dydx$ without a function but i am not sure. Thank you for any help in advance.

calculus integration multivariable-calculus volume

share|cite|improve this question

edited May 27 at 7:26

Boris Borovski

asked May 27 at 7:24

Boris BorovskiBoris Borovski

556

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ...and the right answer is...?
  $endgroup$
  – DonAntonio
  May 27 at 7:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonAntonio I mean the approach i take its not right i get the sarcasm : )
  $endgroup$
  – Boris Borovski
  May 27 at 7:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bo Ok, fine...and then how do you know you're not getting "the right answer"??
  $endgroup$
  – DonAntonio
  May 27 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonAntonio When i evaluate the integral the and i substitute the limits the terms are cancelling out to 0 and the answer is $frac1712-2ln2$. I didn't solve a problem which i need to find just the volume.
  $endgroup$
  – Boris Borovski
  May 27 at 7:44
  

  
  
  
  

add a comment | 

$begingroup$

Hello i am trying to calculate the volume for a double integral but i am having problem with define the integral because it is not given in a pure form. I have $z = xy$, $x+y+z=1$ $z=0$ my approach is to set the function for a integral to be $$int_Dxy$$ and to find the $limits for$ $dy$ i set $z$ to be zero it is also given by definition and i get $y = 1-x$ after that i set both $z$ and $y$ to zero and i get $x = 1$ so i have the followinglimits $$int_0^1 int_0^x-1xy$$ but i am not getting the right answer after evaluating the integral. What confusing me here is that the integral is not given by default here also the other thing that confuses me is i have the same problem but to be solved with triple integral. I am thinking maybe for the volume i just need $dydx$ without a function but i am not sure. Thank you for any help in advance.

calculus integration multivariable-calculus volume

share|cite|improve this question

edited May 27 at 7:26

Boris Borovski

asked May 27 at 7:24

Boris BorovskiBoris Borovski

556

$endgroup$

share|cite|improve this question

edited May 27 at 7:26

Boris Borovski

asked May 27 at 7:24

Boris BorovskiBoris Borovski

556

share|cite|improve this question

edited May 27 at 7:26

Boris Borovski

asked May 27 at 7:24

Boris BorovskiBoris Borovski

556

asked May 27 at 7:24

Boris BorovskiBoris Borovski

556

asked May 27 at 7:24

Boris BorovskiBoris Borovski

556

2 Answers
2

active

oldest

votes

4

$begingroup$

Note that
$$1-x-ygeq xygeq 0quad textin $D_1=leftyin left[0,frac1-x1+xright],xin[0,1]right$$$

and
$$0leq 1-x-yleq xyquad textin $D_2=leftyin left[frac1-x1+x,1-xright],xin[0,1]right$.$$
Therefore, the volume is given by
$$beginalignV&=iint_D_1xy dxdy+iint_D_2(1-x-y) dxdy\
&=int_0^1int_0^frac1-x1+xxy dydx+
int_0^1int_frac1-x1+x^1-x(1-x-y) dydx.
endalign.$$
The final result should be $17/12-2ln(2)$.

share|cite|improve this answer

edited May 27 at 7:59

answered May 27 at 7:46

Robert ZRobert Z

103k1073147

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly the same the OP commented above....
  $endgroup$
  – DonAntonio
  May 27 at 7:48
  

  
  
  
  

add a comment | 

2

$begingroup$

The limits of integration in your $$int_0^1 int_0^x-1xy$$ are not correct.

Note that you have to intersect $$x+y+z=1$$ with $$z=xy$$ which gives you $$ y=frac 1-x1+x$$

Thus the limits of integration are $$int_0^1 int_0^frac 1-x1+xxy dydx$$

share|cite|improve this answer

answered May 27 at 7:37

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert Z I didn't realize i can use $z=xy$ for intersecting it with $x+y+z = 1$,because i was thinking its the function itself. Thank you very much both.
  $endgroup$
  – Boris Borovski
  May 27 at 7:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you also for the help
  $endgroup$
  – Boris Borovski
  May 27 at 7:55
  

  
  
  
  

add a comment | 

Your Answer

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4

$begingroup$

Note that
$$1-x-ygeq xygeq 0quad textin $D_1=leftyin left[0,frac1-x1+xright],xin[0,1]right$$$

and
$$0leq 1-x-yleq xyquad textin $D_2=leftyin left[frac1-x1+x,1-xright],xin[0,1]right$.$$
Therefore, the volume is given by
$$beginalignV&=iint_D_1xy dxdy+iint_D_2(1-x-y) dxdy\
&=int_0^1int_0^frac1-x1+xxy dydx+
int_0^1int_frac1-x1+x^1-x(1-x-y) dydx.
endalign.$$
The final result should be $17/12-2ln(2)$.

share|cite|improve this answer

edited May 27 at 7:59

answered May 27 at 7:46

Robert ZRobert Z

103k1073147

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly the same the OP commented above....
  $endgroup$
  – DonAntonio
  May 27 at 7:48
  

  
  
  
  

add a comment | 

4

$begingroup$

Note that
$$1-x-ygeq xygeq 0quad textin $D_1=leftyin left[0,frac1-x1+xright],xin[0,1]right$$$

and
$$0leq 1-x-yleq xyquad textin $D_2=leftyin left[frac1-x1+x,1-xright],xin[0,1]right$.$$
Therefore, the volume is given by
$$beginalignV&=iint_D_1xy dxdy+iint_D_2(1-x-y) dxdy\
&=int_0^1int_0^frac1-x1+xxy dydx+
int_0^1int_frac1-x1+x^1-x(1-x-y) dydx.
endalign.$$
The final result should be $17/12-2ln(2)$.

share|cite|improve this answer

edited May 27 at 7:59

answered May 27 at 7:46

Robert ZRobert Z

103k1073147

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly the same the OP commented above....
  $endgroup$
  – DonAntonio
  May 27 at 7:48
  

  
  
  
  

add a comment | 

$begingroup$

Note that
$$1-x-ygeq xygeq 0quad textin $D_1=leftyin left[0,frac1-x1+xright],xin[0,1]right$$$

and
$$0leq 1-x-yleq xyquad textin $D_2=leftyin left[frac1-x1+x,1-xright],xin[0,1]right$.$$
Therefore, the volume is given by
$$beginalignV&=iint_D_1xy dxdy+iint_D_2(1-x-y) dxdy\
&=int_0^1int_0^frac1-x1+xxy dydx+
int_0^1int_frac1-x1+x^1-x(1-x-y) dydx.
endalign.$$
The final result should be $17/12-2ln(2)$.

share|cite|improve this answer

edited May 27 at 7:59

answered May 27 at 7:46

Robert ZRobert Z

103k1073147

$endgroup$

share|cite|improve this answer

edited May 27 at 7:59

answered May 27 at 7:46

Robert ZRobert Z

103k1073147

answered May 27 at 7:46

Robert ZRobert Z

103k1073147

answered May 27 at 7:46

Robert ZRobert Z

103k1073147

2

$begingroup$

The limits of integration in your $$int_0^1 int_0^x-1xy$$ are not correct.

Note that you have to intersect $$x+y+z=1$$ with $$z=xy$$ which gives you $$ y=frac 1-x1+x$$

Thus the limits of integration are $$int_0^1 int_0^frac 1-x1+xxy dydx$$

share|cite|improve this answer

answered May 27 at 7:37

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert Z I didn't realize i can use $z=xy$ for intersecting it with $x+y+z = 1$,because i was thinking its the function itself. Thank you very much both.
  $endgroup$
  – Boris Borovski
  May 27 at 7:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you also for the help
  $endgroup$
  – Boris Borovski
  May 27 at 7:55
  

  
  
  
  

add a comment | 

2

$begingroup$

The limits of integration in your $$int_0^1 int_0^x-1xy$$ are not correct.

Note that you have to intersect $$x+y+z=1$$ with $$z=xy$$ which gives you $$ y=frac 1-x1+x$$

Thus the limits of integration are $$int_0^1 int_0^frac 1-x1+xxy dydx$$

share|cite|improve this answer

answered May 27 at 7:37

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert Z I didn't realize i can use $z=xy$ for intersecting it with $x+y+z = 1$,because i was thinking its the function itself. Thank you very much both.
  $endgroup$
  – Boris Borovski
  May 27 at 7:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you also for the help
  $endgroup$
  – Boris Borovski
  May 27 at 7:55
  

  
  
  
  

add a comment | 

$begingroup$

The limits of integration in your $$int_0^1 int_0^x-1xy$$ are not correct.

Note that you have to intersect $$x+y+z=1$$ with $$z=xy$$ which gives you $$ y=frac 1-x1+x$$

Thus the limits of integration are $$int_0^1 int_0^frac 1-x1+xxy dydx$$

share|cite|improve this answer

answered May 27 at 7:37

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

$endgroup$

share|cite|improve this answer

answered May 27 at 7:37

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

answered May 27 at 7:37

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

answered May 27 at 7:37

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163