Fdtxjr

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List elements digit difference sort

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How should I solve this integral with changing parameters?

How do I calculate a double integral like $int_mathbfDint e^fracx-yx+y dx dy$?$int_D e^fracx-yx+ydxdy$ Where is my mistake?$-iint_A(y+x),dA$ EvaluatingChanging Variables in double integralUsing change of variables, solve the integral and show the domain obtained by the change.Integration w/ Change of VariablesChange of Variables for double integrationHow to find an equation that goes through two points, with a definite integral.How to solve this triple integral on $[-1,1]^3$?How to solve the integral $IR_0int_0^sqrtR^2-R_0^2(R_0^2+x^2)^-frac32dx$?How to solve this beautiful integral equationhow to solve $int_-infty^t sin(omega_o(t-t'))(1+tanh(fract'tau))dt'$?

5

$begingroup$

I can't solve this. How should I proceed?

$$iint_De^largefracy-xy+xmathrm dxmathrm dy$$

$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-frac12$ but I have problem finding the range of $u$ and $v$ to calculate the integral.

definite-integrals

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edited yesterday

Max

564317

asked 2 days ago

khoshrangkhoshrang

596

New contributor

khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558
  $endgroup$
  – Jean Marie
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks !
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  

add a comment | 

5

$begingroup$

I can't solve this. How should I proceed?

$$iint_De^largefracy-xy+xmathrm dxmathrm dy$$

$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-frac12$ but I have problem finding the range of $u$ and $v$ to calculate the integral.

definite-integrals

share|cite|improve this question

edited yesterday

Max

564317

asked 2 days ago

khoshrangkhoshrang

596

New contributor

khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558
  $endgroup$
  – Jean Marie
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks !
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

I can't solve this. How should I proceed?

$$iint_De^largefracy-xy+xmathrm dxmathrm dy$$

$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-frac12$ but I have problem finding the range of $u$ and $v$ to calculate the integral.

definite-integrals

share|cite|improve this question

edited yesterday

Max

564317

asked 2 days ago

khoshrangkhoshrang

596

New contributor

khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

Max

564317

asked 2 days ago

khoshrangkhoshrang

596

New contributor

khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

Max

564317

asked 2 days ago

khoshrangkhoshrang

596

New contributor

khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

Max

564317

edited yesterday

Max

564317

asked 2 days ago

khoshrangkhoshrang

596

New contributor

khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

khoshrangkhoshrang

596

3 Answers
3

active

oldest

votes

8

$begingroup$

Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.

share|cite|improve this answer

answered 2 days ago

AndreiAndrei

13.1k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
  $endgroup$
  – Andrei
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
  $endgroup$
  – Andrei
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Your domain is $$D=(x,y):0<x<2,,,0<y<2,,,x+y<2$$

Using change of variables $(x,y)to(u,v)$ with $$u=fracx-yx+y,,,v=x+y$$

The region is now $$R=(u,v):-1<u<1,,0<v<2$$

Therefore,

beginalign
iint_D expleft(-fracx-yx+yright),mathrmdx,mathrmdy&=frac12iint_R ve^-u,mathrmdu,mathrmdv
\\&=frac12int_0^2v,mathrmdvint_-1^1 e^-u,mathrmdu
endalign

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

StubbornAtomStubbornAtom

6,16811339

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Finally the $v$ appears, but with no explanation at all?
  $endgroup$
  – NickD
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NickD I have left the calculation of the jacobian to the OP.
  $endgroup$
  – StubbornAtom
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's fine, but you should at least mention it, if only for the sake of future readers.
  $endgroup$
  – NickD
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @khoshrang To see the bounds of $u$ and $v$, replace the string of inequalities in $D$ with $u$ and $v$. (So first of all, solve for $x,y$ in terms of $u,v$ ). And $v/2$ is the jacobian of course, no surprise that it appears at the end.
  $endgroup$
  – StubbornAtom
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
  $endgroup$
  – khoshrang
  2 days ago
  
  

  
  
  
  

 | 
show 3 more comments

0

$begingroup$

Computer algebra gives (for the general case):

$$e left(x^2 textEileft(-frac2 xx+yright)-fracy^2 textEileft(frac2
yx+yright)e^2right)+frac12 e^1-frac2 xx+y (x+y)^2$$

over your specified region:

$$e-frac1e$$

share|cite|improve this answer

answered 2 days ago

David G. StorkDavid G. Stork

11.1k41432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
  $endgroup$
  – khoshrang
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (-1) because this is not an answer to the question. See the other excellent responses!
  $endgroup$
  – Jimmy Sabater
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
  $endgroup$
  – David G. Stork
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
  $endgroup$
  – Mikey Spivak
  2 days ago
  

  
  
  
  

add a comment | 

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8

$begingroup$

Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.

share|cite|improve this answer

answered 2 days ago

AndreiAndrei

13.1k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
  $endgroup$
  – Andrei
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
  $endgroup$
  – Andrei
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  

add a comment | 

8

$begingroup$

Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.

share|cite|improve this answer

answered 2 days ago

AndreiAndrei

13.1k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
  $endgroup$
  – Andrei
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
  $endgroup$
  – Andrei
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
  $endgroup$
  – khoshrang
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.

share|cite|improve this answer

answered 2 days ago

AndreiAndrei

13.1k21230

$endgroup$

share|cite|improve this answer

answered 2 days ago

AndreiAndrei

13.1k21230

answered 2 days ago

AndreiAndrei

13.1k21230

answered 2 days ago

AndreiAndrei

13.1k21230

6

$begingroup$

Your domain is $$D=(x,y):0<x<2,,,0<y<2,,,x+y<2$$

Using change of variables $(x,y)to(u,v)$ with $$u=fracx-yx+y,,,v=x+y$$

The region is now $$R=(u,v):-1<u<1,,0<v<2$$

Therefore,

beginalign
iint_D expleft(-fracx-yx+yright),mathrmdx,mathrmdy&=frac12iint_R ve^-u,mathrmdu,mathrmdv
\\&=frac12int_0^2v,mathrmdvint_-1^1 e^-u,mathrmdu
endalign

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

StubbornAtomStubbornAtom

6,16811339

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Finally the $v$ appears, but with no explanation at all?
  $endgroup$
  – NickD
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NickD I have left the calculation of the jacobian to the OP.
  $endgroup$
  – StubbornAtom
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's fine, but you should at least mention it, if only for the sake of future readers.
  $endgroup$
  – NickD
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @khoshrang To see the bounds of $u$ and $v$, replace the string of inequalities in $D$ with $u$ and $v$. (So first of all, solve for $x,y$ in terms of $u,v$ ). And $v/2$ is the jacobian of course, no surprise that it appears at the end.
  $endgroup$
  – StubbornAtom
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
  $endgroup$
  – khoshrang
  2 days ago
  
  

  
  
  
  

 | 
show 3 more comments

6

$begingroup$

Your domain is $$D=(x,y):0<x<2,,,0<y<2,,,x+y<2$$

Using change of variables $(x,y)to(u,v)$ with $$u=fracx-yx+y,,,v=x+y$$

The region is now $$R=(u,v):-1<u<1,,0<v<2$$

Therefore,

beginalign
iint_D expleft(-fracx-yx+yright),mathrmdx,mathrmdy&=frac12iint_R ve^-u,mathrmdu,mathrmdv
\\&=frac12int_0^2v,mathrmdvint_-1^1 e^-u,mathrmdu
endalign

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

StubbornAtomStubbornAtom

6,16811339

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Finally the $v$ appears, but with no explanation at all?
  $endgroup$
  – NickD
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NickD I have left the calculation of the jacobian to the OP.
  $endgroup$
  – StubbornAtom
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's fine, but you should at least mention it, if only for the sake of future readers.
  $endgroup$
  – NickD
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  @khoshrang To see the bounds of $u$ and $v$, replace the string of inequalities in $D$ with $u$ and $v$. (So first of all, solve for $x,y$ in terms of $u,v$ ). And $v/2$ is the jacobian of course, no surprise that it appears at the end.
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  – StubbornAtom
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  thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
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  – khoshrang
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show 3 more comments

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Your domain is $$D=(x,y):0<x<2,,,0<y<2,,,x+y<2$$

Using change of variables $(x,y)to(u,v)$ with $$u=fracx-yx+y,,,v=x+y$$

The region is now $$R=(u,v):-1<u<1,,0<v<2$$

Therefore,

beginalign
iint_D expleft(-fracx-yx+yright),mathrmdx,mathrmdy&=frac12iint_R ve^-u,mathrmdu,mathrmdv
\\&=frac12int_0^2v,mathrmdvint_-1^1 e^-u,mathrmdu
endalign

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

StubbornAtomStubbornAtom

6,16811339

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share|cite|improve this answer

edited 2 days ago

answered 2 days ago

StubbornAtomStubbornAtom

6,16811339

answered 2 days ago

StubbornAtomStubbornAtom

6,16811339

answered 2 days ago

StubbornAtomStubbornAtom

6,16811339

0

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Computer algebra gives (for the general case):

$$e left(x^2 textEileft(-frac2 xx+yright)-fracy^2 textEileft(frac2
yx+yright)e^2right)+frac12 e^1-frac2 xx+y (x+y)^2$$

over your specified region:

$$e-frac1e$$

share|cite|improve this answer

answered 2 days ago

David G. StorkDavid G. Stork

11.1k41432

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  yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
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  – khoshrang
  2 days ago
  
  

  
  
  
  


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  (-1) because this is not an answer to the question. See the other excellent responses!
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  – Jimmy Sabater
  2 days ago
  

  
  
  
  


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  @JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
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  – David G. Stork
  2 days ago
  

  
  
  
  


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  The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
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  – Mikey Spivak
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Computer algebra gives (for the general case):

$$e left(x^2 textEileft(-frac2 xx+yright)-fracy^2 textEileft(frac2
yx+yright)e^2right)+frac12 e^1-frac2 xx+y (x+y)^2$$

over your specified region:

$$e-frac1e$$

share|cite|improve this answer

answered 2 days ago

David G. StorkDavid G. Stork

11.1k41432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
  $endgroup$
  – khoshrang
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (-1) because this is not an answer to the question. See the other excellent responses!
  $endgroup$
  – Jimmy Sabater
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
  $endgroup$
  – David G. Stork
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
  $endgroup$
  – Mikey Spivak
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Computer algebra gives (for the general case):

$$e left(x^2 textEileft(-frac2 xx+yright)-fracy^2 textEileft(frac2
yx+yright)e^2right)+frac12 e^1-frac2 xx+y (x+y)^2$$

over your specified region:

$$e-frac1e$$

share|cite|improve this answer

answered 2 days ago

David G. StorkDavid G. Stork

11.1k41432

$endgroup$

share|cite|improve this answer

answered 2 days ago

David G. StorkDavid G. Stork

11.1k41432

answered 2 days ago

David G. StorkDavid G. Stork

11.1k41432

answered 2 days ago

David G. StorkDavid G. Stork

11.1k41432