$iint e^{x-y}$ over the triangle with vertices at $(0,0),(1,3),(2,2)$
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$displaystyleiint e^{x-y}$ over the triangle with vertices at $(0,0),(1,3),(2,2)$
I tried the following change of variables. let $u=x-y$, $v=3x-y$. In $(u,v)$, we get the triangle with vertices at $(0,0),(-2,0),(0,4)$.The Jacobian i calculated is $1/2$. I tried integrating $(1/2)displaystyleiint e^u$ over this new region and got $2/e^2$. The answer provided $1+1/e^2$, I'm not too sure why my change of variables is not working. Any help would be greatly appreciated
multivariable-calculus
edited Nov 23 at 16:28
Brahadeesh
5,78441957
asked Nov 23 at 15:43
mmmmo
1027
add a comment |
up vote
1
down vote
favorite
$displaystyleiint e^{x-y}$ over the triangle with vertices at $(0,0),(1,3),(2,2)$
I tried the following change of variables. let $u=x-y$, $v=3x-y$. In $(u,v)$, we get the triangle with vertices at $(0,0),(-2,0),(0,4)$.The Jacobian i calculated is $1/2$. I tried integrating $(1/2)displaystyleiint e^u$ over this new region and got $2/e^2$. The answer provided $1+1/e^2$, I'm not too sure why my change of variables is not working. Any help would be greatly appreciated
multivariable-calculus
edited Nov 23 at 16:28
Brahadeesh
5,78441957
asked Nov 23 at 15:43
mmmmo
1027
For such simple form of the integrand, changing variables is not necessary. Drawing a sketch of the region is enough to find the limits of integration.
– StubbornAtom
Nov 23 at 17:45
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$displaystyleiint e^{x-y}$ over the triangle with vertices at $(0,0),(1,3),(2,2)$
I tried the following change of variables. let $u=x-y$, $v=3x-y$. In $(u,v)$, we get the triangle with vertices at $(0,0),(-2,0),(0,4)$.The Jacobian i calculated is $1/2$. I tried integrating $(1/2)displaystyleiint e^u$ over this new region and got $2/e^2$. The answer provided $1+1/e^2$, I'm not too sure why my change of variables is not working. Any help would be greatly appreciated
multivariable-calculus
edited Nov 23 at 16:28
Brahadeesh
5,78441957
asked Nov 23 at 15:43
mmmmo
1027
$displaystyleiint e^{x-y}$ over the triangle with vertices at $(0,0),(1,3),(2,2)$
I tried the following change of variables. let $u=x-y$, $v=3x-y$. In $(u,v)$, we get the triangle with vertices at $(0,0),(-2,0),(0,4)$.The Jacobian i calculated is $1/2$. I tried integrating $(1/2)displaystyleiint e^u$ over this new region and got $2/e^2$. The answer provided $1+1/e^2$, I'm not too sure why my change of variables is not working. Any help would be greatly appreciated
multivariable-calculus
multivariable-calculus
edited Nov 23 at 16:28
Brahadeesh
5,78441957
asked Nov 23 at 15:43
mmmmo
1027
edited Nov 23 at 16:28
Brahadeesh
5,78441957
asked Nov 23 at 15:43
mmmmo
1027
edited Nov 23 at 16:28
Brahadeesh
5,78441957
edited Nov 23 at 16:28
Brahadeesh
5,78441957
edited Nov 23 at 16:28
Brahadeesh
5,78441957
5,78441957
asked Nov 23 at 15:43
mmmmo
1027
asked Nov 23 at 15:43
mmmmo
1027
asked Nov 23 at 15:43
mmmmo
1027
1027
For such simple form of the integrand, changing variables is not necessary. Drawing a sketch of the region is enough to find the limits of integration.
– StubbornAtom
Nov 23 at 17:45
add a comment |
For such simple form of the integrand, changing variables is not necessary. Drawing a sketch of the region is enough to find the limits of integration.
– StubbornAtom
Nov 23 at 17:45
For such simple form of the integrand, changing variables is not necessary. Drawing a sketch of the region is enough to find the limits of integration.
– StubbornAtom
Nov 23 at 17:45
For such simple form of the integrand, changing variables is not necessary. Drawing a sketch of the region is enough to find the limits of integration.
– StubbornAtom
Nov 23 at 17:45
add a comment |
1 Answer
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First of all, I assume you mean the region $R$ enclosed by the triangle, not the triangle itself. Otherwise the integral is just zero.
Instead of bothering with Jacobians, which I personally find super annoying, it may be best simply to break it up into two pieces. Thus we have
$$iintlimits_R e^{x-y};dA=int_0^1int_x^{3x}e^{x-y};dy;dx+int_1^2int_x^{4-x}e^{x-y};dy;dx$$
$$=left(frac{1}{2}+frac{1}{2e^2}right)+left(frac{1}{2}+frac{1}{2e^2}right)=1+frac{1}{e^2}.$$
answered Nov 23 at 16:26
Ben W
1,148510
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
First of all, I assume you mean the region $R$ enclosed by the triangle, not the triangle itself. Otherwise the integral is just zero.
Instead of bothering with Jacobians, which I personally find super annoying, it may be best simply to break it up into two pieces. Thus we have
$$iintlimits_R e^{x-y};dA=int_0^1int_x^{3x}e^{x-y};dy;dx+int_1^2int_x^{4-x}e^{x-y};dy;dx$$
$$=left(frac{1}{2}+frac{1}{2e^2}right)+left(frac{1}{2}+frac{1}{2e^2}right)=1+frac{1}{e^2}.$$
answered Nov 23 at 16:26
Ben W
1,148510
add a comment |
up vote
2
down vote
First of all, I assume you mean the region $R$ enclosed by the triangle, not the triangle itself. Otherwise the integral is just zero.
Instead of bothering with Jacobians, which I personally find super annoying, it may be best simply to break it up into two pieces. Thus we have
$$iintlimits_R e^{x-y};dA=int_0^1int_x^{3x}e^{x-y};dy;dx+int_1^2int_x^{4-x}e^{x-y};dy;dx$$
$$=left(frac{1}{2}+frac{1}{2e^2}right)+left(frac{1}{2}+frac{1}{2e^2}right)=1+frac{1}{e^2}.$$
answered Nov 23 at 16:26
Ben W
1,148510
add a comment |
up vote
2
down vote
up vote
2
down vote
First of all, I assume you mean the region $R$ enclosed by the triangle, not the triangle itself. Otherwise the integral is just zero.
Instead of bothering with Jacobians, which I personally find super annoying, it may be best simply to break it up into two pieces. Thus we have
$$iintlimits_R e^{x-y};dA=int_0^1int_x^{3x}e^{x-y};dy;dx+int_1^2int_x^{4-x}e^{x-y};dy;dx$$
$$=left(frac{1}{2}+frac{1}{2e^2}right)+left(frac{1}{2}+frac{1}{2e^2}right)=1+frac{1}{e^2}.$$
answered Nov 23 at 16:26
Ben W
1,148510
First of all, I assume you mean the region $R$ enclosed by the triangle, not the triangle itself. Otherwise the integral is just zero.
Instead of bothering with Jacobians, which I personally find super annoying, it may be best simply to break it up into two pieces. Thus we have
$$iintlimits_R e^{x-y};dA=int_0^1int_x^{3x}e^{x-y};dy;dx+int_1^2int_x^{4-x}e^{x-y};dy;dx$$
$$=left(frac{1}{2}+frac{1}{2e^2}right)+left(frac{1}{2}+frac{1}{2e^2}right)=1+frac{1}{e^2}.$$
answered Nov 23 at 16:26
Ben W
1,148510
answered Nov 23 at 16:26
Ben W
1,148510
answered Nov 23 at 16:26
Ben W
1,148510
answered Nov 23 at 16:26
Ben W
1,148510
1,148510
add a comment |
add a comment |
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For such simple form of the integrand, changing variables is not necessary. Drawing a sketch of the region is enough to find the limits of integration.
– StubbornAtom
Nov 23 at 17:45