Mixing
Evaluate $intint_Ge^{yover x+y}dxdy$
$begingroup$
Evaluate $intint_Ge^{yover x+y}dxdy$ where G is the triangle enclosed by $x+y=1$, x axis and y axis.
My attempt:
let $u=x+y$, $v=y$, $|J|=1$
$$int_0^1int_0^{-x+1}e^{yover x+y}dydx=int_0^1int_0^1e^{vover u}dvdu$$
But still I couldn't figure out this.
calculus
asked Jan 27 at 18:40
Yibei HeYibei He
3139
$endgroup$
add a comment |
$begingroup$
Evaluate $intint_Ge^{yover x+y}dxdy$ where G is the triangle enclosed by $x+y=1$, x axis and y axis.
My attempt:
let $u=x+y$, $v=y$, $|J|=1$
$$int_0^1int_0^{-x+1}e^{yover x+y}dydx=int_0^1int_0^1e^{vover u}dvdu$$
But still I couldn't figure out this.
calculus
asked Jan 27 at 18:40
Yibei HeYibei He
3139
$endgroup$
$begingroup$
In this triangle, $u ge v$ so you'd want the $v$ integral to go from $0$ to $u$.
$endgroup$
– Robert Israel
Jan 27 at 18:54
$begingroup$
Yes, I did miss that, thank you!
$endgroup$
– Yibei He
Jan 27 at 19:00
add a comment |
$begingroup$
Evaluate $intint_Ge^{yover x+y}dxdy$ where G is the triangle enclosed by $x+y=1$, x axis and y axis.
My attempt:
let $u=x+y$, $v=y$, $|J|=1$
$$int_0^1int_0^{-x+1}e^{yover x+y}dydx=int_0^1int_0^1e^{vover u}dvdu$$
But still I couldn't figure out this.
calculus
asked Jan 27 at 18:40
Yibei HeYibei He
3139
$endgroup$
Evaluate $intint_Ge^{yover x+y}dxdy$ where G is the triangle enclosed by $x+y=1$, x axis and y axis.
My attempt:
let $u=x+y$, $v=y$, $|J|=1$
$$int_0^1int_0^{-x+1}e^{yover x+y}dydx=int_0^1int_0^1e^{vover u}dvdu$$
But still I couldn't figure out this.
calculus
calculus
asked Jan 27 at 18:40
Yibei HeYibei He
3139
asked Jan 27 at 18:40
Yibei HeYibei He
3139
asked Jan 27 at 18:40
Yibei HeYibei He
3139
asked Jan 27 at 18:40
Yibei HeYibei He
3139
asked Jan 27 at 18:40
Yibei HeYibei He
3139
3139
$begingroup$
In this triangle, $u ge v$ so you'd want the $v$ integral to go from $0$ to $u$.
$endgroup$
– Robert Israel
Jan 27 at 18:54
$begingroup$
Yes, I did miss that, thank you!
$endgroup$
– Yibei He
Jan 27 at 19:00
add a comment |
$begingroup$
In this triangle, $u ge v$ so you'd want the $v$ integral to go from $0$ to $u$.
$endgroup$
– Robert Israel
Jan 27 at 18:54
$begingroup$
Yes, I did miss that, thank you!
$endgroup$
– Yibei He
Jan 27 at 19:00
$begingroup$
In this triangle, $u ge v$ so you'd want the $v$ integral to go from $0$ to $u$.
$endgroup$
– Robert Israel
Jan 27 at 18:54
$begingroup$
In this triangle, $u ge v$ so you'd want the $v$ integral to go from $0$ to $u$.
$endgroup$
– Robert Israel
Jan 27 at 18:54
$begingroup$
Yes, I did miss that, thank you!
$endgroup$
– Yibei He
Jan 27 at 19:00
$begingroup$
Yes, I did miss that, thank you!
$endgroup$
– Yibei He
Jan 27 at 19:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the substitution
$$x=rcos^2phi,quad y=rsin^2phi,
$$
with $|J|=rsin2phi $, which results in
$$begin{array}{}
intint_Ge^{yover x+y}dxdy=int_0^1 drint_0^{pi/2}e^{sin^2phi}rsin2phi; dphi\
stackrel{sin^2phimapsto u}=int_0^1 rdrint_0^{1}e^udu=frac {e-1}{2}.
end{array}
$$
answered Jan 27 at 20:50
useruser
5,86511031
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the substitution
$$x=rcos^2phi,quad y=rsin^2phi,
$$
with $|J|=rsin2phi $, which results in
$$begin{array}{}
intint_Ge^{yover x+y}dxdy=int_0^1 drint_0^{pi/2}e^{sin^2phi}rsin2phi; dphi\
stackrel{sin^2phimapsto u}=int_0^1 rdrint_0^{1}e^udu=frac {e-1}{2}.
end{array}
$$
answered Jan 27 at 20:50
useruser
5,86511031
$endgroup$
add a comment |
$begingroup$
Consider the substitution
$$x=rcos^2phi,quad y=rsin^2phi,
$$
with $|J|=rsin2phi $, which results in
$$begin{array}{}
intint_Ge^{yover x+y}dxdy=int_0^1 drint_0^{pi/2}e^{sin^2phi}rsin2phi; dphi\
stackrel{sin^2phimapsto u}=int_0^1 rdrint_0^{1}e^udu=frac {e-1}{2}.
end{array}
$$
answered Jan 27 at 20:50
useruser
5,86511031
$endgroup$
add a comment |
$begingroup$
Consider the substitution
$$x=rcos^2phi,quad y=rsin^2phi,
$$
with $|J|=rsin2phi $, which results in
$$begin{array}{}
intint_Ge^{yover x+y}dxdy=int_0^1 drint_0^{pi/2}e^{sin^2phi}rsin2phi; dphi\
stackrel{sin^2phimapsto u}=int_0^1 rdrint_0^{1}e^udu=frac {e-1}{2}.
end{array}
$$
answered Jan 27 at 20:50
useruser
5,86511031
$endgroup$
Consider the substitution
$$x=rcos^2phi,quad y=rsin^2phi,
$$
with $|J|=rsin2phi $, which results in
$$begin{array}{}
intint_Ge^{yover x+y}dxdy=int_0^1 drint_0^{pi/2}e^{sin^2phi}rsin2phi; dphi\
stackrel{sin^2phimapsto u}=int_0^1 rdrint_0^{1}e^udu=frac {e-1}{2}.
end{array}
$$
answered Jan 27 at 20:50
useruser
5,86511031
edited Jan 28 at 21:25
answered Jan 27 at 20:50
useruser
5,86511031
answered Jan 27 at 20:50
useruser
5,86511031
answered Jan 27 at 20:50
useruser
5,86511031
5,86511031
add a comment |
add a comment |
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$begingroup$
In this triangle, $u ge v$ so you'd want the $v$ integral to go from $0$ to $u$.
$endgroup$
– Robert Israel
Jan 27 at 18:54
$begingroup$
Yes, I did miss that, thank you!
$endgroup$
– Yibei He
Jan 27 at 19:00