Have trouble finding the bounds to this triple integration?
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I have been trying to solve the problem below, but I have kept arriving at the same answer. I think I set up the bounds incorrectly, but I do not have a sure way of knowing:
Find the volume of solid bounded by 3 coordinate planes, bounded above by plane x+y+z=2 and bounded below by z=x+y.
The integration problem is then
$V = displaystyle int_x int_y int_z=x+y^2-x-y 1dzdydx$, which can be simplified to
$V = displaystyle int_x int_y (2-x-y-(x+y))dydx$, or
$V = displaystyle int_x int_y (2-2x-2y)dydx$.
I calculated the bounds for x to be $0 leq x leq 2$ and for y to be
$0 leq y leq 2-x$. Thus, my construction of the integral becomes
$V = displaystyle int_x=0^2 int_y=0^2-x (2-2x-2y)dydx$ which evaluates to $-frac43$. This not correct. The integral is actually $frac13$.
integration multivariable-calculus
asked Aug 16 at 1:50
Ronny
1
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I have been trying to solve the problem below, but I have kept arriving at the same answer. I think I set up the bounds incorrectly, but I do not have a sure way of knowing:
Find the volume of solid bounded by 3 coordinate planes, bounded above by plane x+y+z=2 and bounded below by z=x+y.
The integration problem is then
$V = displaystyle int_x int_y int_z=x+y^2-x-y 1dzdydx$, which can be simplified to
$V = displaystyle int_x int_y (2-x-y-(x+y))dydx$, or
$V = displaystyle int_x int_y (2-2x-2y)dydx$.
I calculated the bounds for x to be $0 leq x leq 2$ and for y to be
$0 leq y leq 2-x$. Thus, my construction of the integral becomes
$V = displaystyle int_x=0^2 int_y=0^2-x (2-2x-2y)dydx$ which evaluates to $-frac43$. This not correct. The integral is actually $frac13$.
integration multivariable-calculus
asked Aug 16 at 1:50
Ronny
1
should be $x=0$ to $x=1$ and $y$ from $0$ to $1-x$
– Lozenges
Aug 16 at 3:30
Much appreciated.
– Ronny
Aug 16 at 3:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have been trying to solve the problem below, but I have kept arriving at the same answer. I think I set up the bounds incorrectly, but I do not have a sure way of knowing:
Find the volume of solid bounded by 3 coordinate planes, bounded above by plane x+y+z=2 and bounded below by z=x+y.
The integration problem is then
$V = displaystyle int_x int_y int_z=x+y^2-x-y 1dzdydx$, which can be simplified to
$V = displaystyle int_x int_y (2-x-y-(x+y))dydx$, or
$V = displaystyle int_x int_y (2-2x-2y)dydx$.
I calculated the bounds for x to be $0 leq x leq 2$ and for y to be
$0 leq y leq 2-x$. Thus, my construction of the integral becomes
$V = displaystyle int_x=0^2 int_y=0^2-x (2-2x-2y)dydx$ which evaluates to $-frac43$. This not correct. The integral is actually $frac13$.
integration multivariable-calculus
asked Aug 16 at 1:50
Ronny
1
I have been trying to solve the problem below, but I have kept arriving at the same answer. I think I set up the bounds incorrectly, but I do not have a sure way of knowing:
Find the volume of solid bounded by 3 coordinate planes, bounded above by plane x+y+z=2 and bounded below by z=x+y.
The integration problem is then
$V = displaystyle int_x int_y int_z=x+y^2-x-y 1dzdydx$, which can be simplified to
$V = displaystyle int_x int_y (2-x-y-(x+y))dydx$, or
$V = displaystyle int_x int_y (2-2x-2y)dydx$.
I calculated the bounds for x to be $0 leq x leq 2$ and for y to be
$0 leq y leq 2-x$. Thus, my construction of the integral becomes
$V = displaystyle int_x=0^2 int_y=0^2-x (2-2x-2y)dydx$ which evaluates to $-frac43$. This not correct. The integral is actually $frac13$.
integration multivariable-calculus
asked Aug 16 at 1:50
Ronny
1
asked Aug 16 at 1:50
Ronny
1
asked Aug 16 at 1:50
Ronny
1
asked Aug 16 at 1:50
Ronny
1
1
should be $x=0$ to $x=1$ and $y$ from $0$ to $1-x$
– Lozenges
Aug 16 at 3:30
Much appreciated.
– Ronny
Aug 16 at 3:40
add a comment |Â
should be $x=0$ to $x=1$ and $y$ from $0$ to $1-x$
– Lozenges
Aug 16 at 3:30
Much appreciated.
– Ronny
Aug 16 at 3:40
should be $x=0$ to $x=1$ and $y$ from $0$ to $1-x$
– Lozenges
Aug 16 at 3:30
should be $x=0$ to $x=1$ and $y$ from $0$ to $1-x$
– Lozenges
Aug 16 at 3:30
Much appreciated.
– Ronny
Aug 16 at 3:40
Much appreciated.
– Ronny
Aug 16 at 3:40
add a comment |Â
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should be $x=0$ to $x=1$ and $y$ from $0$ to $1-x$
– Lozenges
Aug 16 at 3:30
Much appreciated.
– Ronny
Aug 16 at 3:40