Evaluate $intint_Ge^yover x+ydxdy$evaluating this difficult integralSurface Integral over a Vector Field questionEvaluate$int_-2^2int_y^2-3^5-y^2dxdy$Evaluate $int_-2^2int_y^2-3^5-y^2dxdy$Explain why you could have have predicted the result… By inspection: $int^2_0 int^4-x^2_0 x^3 dydx = frac163 units^3 $Evaluate the double integral boundedEvaluate the double integral $intint_R(x^2-2y) $ $dxdy$Changing integration order of $int_0^1int_e^y-1^e^y f(x,y)dxdy$$iint f(x)g(y)dxdy =int f(x)dx int g(y)dy $ Why?finding area using double integrals
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Evaluate $intint_Ge^yover x+ydxdy$
evaluating this difficult integralSurface Integral over a Vector Field questionEvaluate$int_-2^2int_y^2-3^5-y^2dxdy$Evaluate $int_-2^2int_y^2-3^5-y^2dxdy$Explain why you could have have predicted the result… By inspection: $int^2_0 int^4-x^2_0 x^3 dydx = frac163 units^3 $Evaluate the double integral boundedEvaluate the double integral $intint_R(x^2-2y) $ $dxdy$Changing integration order of $int_0^1int_e^y-1^e^y f(x,y)dxdy$$iint f(x)g(y)dxdy =int f(x)dx int g(y)dy $ Why?finding area using double integrals
$begingroup$
Evaluate $intint_Ge^yover x+ydxdy$ where G is the triangle enclosed by $x+y=1$, x axis and y axis.
My attempt:
let $u=x+y$, $v=y$, $|J|=1$
$$int_0^1int_0^-x+1e^yover x+ydydx=int_0^1int_0^1e^vover udvdu$$
But still I couldn't figure out this.
calculus
$endgroup$
add a comment |
$begingroup$
Evaluate $intint_Ge^yover x+ydxdy$ where G is the triangle enclosed by $x+y=1$, x axis and y axis.
My attempt:
let $u=x+y$, $v=y$, $|J|=1$
$$int_0^1int_0^-x+1e^yover x+ydydx=int_0^1int_0^1e^vover udvdu$$
But still I couldn't figure out this.
calculus
$endgroup$
$begingroup$
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 $intint_Ge^yover x+ydxdy$ where G is the triangle enclosed by $x+y=1$, x axis and y axis.
My attempt:
let $u=x+y$, $v=y$, $|J|=1$
$$int_0^1int_0^-x+1e^yover x+ydydx=int_0^1int_0^1e^vover udvdu$$
But still I couldn't figure out this.
calculus
$endgroup$
Evaluate $intint_Ge^yover x+ydxdy$ where G is the triangle enclosed by $x+y=1$, x axis and y axis.
My attempt:
let $u=x+y$, $v=y$, $|J|=1$
$$int_0^1int_0^-x+1e^yover x+ydydx=int_0^1int_0^1e^vover udvdu$$
But still I couldn't figure out this.
calculus
calculus
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
$$
intint_Ge^yover x+ydxdy=int_0^1 drint_0^pi/2e^sin^2phirsin2phi; dphi\
stackrelsin^2phimapsto u=int_0^1 rdrint_0^1e^udu=frac e-12.
$$
$endgroup$
add a comment |
Your Answer
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1 Answer
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active
oldest
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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
$$
intint_Ge^yover x+ydxdy=int_0^1 drint_0^pi/2e^sin^2phirsin2phi; dphi\
stackrelsin^2phimapsto u=int_0^1 rdrint_0^1e^udu=frac e-12.
$$
$endgroup$
add a comment |
$begingroup$
Consider the substitution
$$x=rcos^2phi,quad y=rsin^2phi,
$$
with $|J|=rsin2phi $, which results in
$$
intint_Ge^yover x+ydxdy=int_0^1 drint_0^pi/2e^sin^2phirsin2phi; dphi\
stackrelsin^2phimapsto u=int_0^1 rdrint_0^1e^udu=frac e-12.
$$
$endgroup$
add a comment |
$begingroup$
Consider the substitution
$$x=rcos^2phi,quad y=rsin^2phi,
$$
with $|J|=rsin2phi $, which results in
$$
intint_Ge^yover x+ydxdy=int_0^1 drint_0^pi/2e^sin^2phirsin2phi; dphi\
stackrelsin^2phimapsto u=int_0^1 rdrint_0^1e^udu=frac e-12.
$$
$endgroup$
Consider the substitution
$$x=rcos^2phi,quad y=rsin^2phi,
$$
with $|J|=rsin2phi $, which results in
$$
intint_Ge^yover x+ydxdy=int_0^1 drint_0^pi/2e^sin^2phirsin2phi; dphi\
stackrelsin^2phimapsto u=int_0^1 rdrint_0^1e^udu=frac e-12.
$$
edited Mar 30 at 0:21
answered Jan 27 at 20:50
useruser
6,34611031
6,34611031
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