Triple Integral bounds question
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For the solid bounded by $x=0 , x=2, z=y, z=y-1, z=0, z=4$ I am looking at the $yz-plane$ and setting up a triple integral. I get $int_{0}^{4} int_{z+1}^{z} int_{0}^{2} dxdydz$ However, I was wondering if I did this correctly because $z+1$ is greater than $z$, but when I drew the projection on the yz plane, it looked like z=y-1 is under z=y. Any help appreciated.
calculus integration
asked Nov 20 at 21:04
darklord0530
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For the solid bounded by $x=0 , x=2, z=y, z=y-1, z=0, z=4$ I am looking at the $yz-plane$ and setting up a triple integral. I get $int_{0}^{4} int_{z+1}^{z} int_{0}^{2} dxdydz$ However, I was wondering if I did this correctly because $z+1$ is greater than $z$, but when I drew the projection on the yz plane, it looked like z=y-1 is under z=y. Any help appreciated.
calculus integration
asked Nov 20 at 21:04
darklord0530
613
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For the solid bounded by $x=0 , x=2, z=y, z=y-1, z=0, z=4$ I am looking at the $yz-plane$ and setting up a triple integral. I get $int_{0}^{4} int_{z+1}^{z} int_{0}^{2} dxdydz$ However, I was wondering if I did this correctly because $z+1$ is greater than $z$, but when I drew the projection on the yz plane, it looked like z=y-1 is under z=y. Any help appreciated.
calculus integration
asked Nov 20 at 21:04
darklord0530
613
For the solid bounded by $x=0 , x=2, z=y, z=y-1, z=0, z=4$ I am looking at the $yz-plane$ and setting up a triple integral. I get $int_{0}^{4} int_{z+1}^{z} int_{0}^{2} dxdydz$ However, I was wondering if I did this correctly because $z+1$ is greater than $z$, but when I drew the projection on the yz plane, it looked like z=y-1 is under z=y. Any help appreciated.
calculus integration
calculus integration
asked Nov 20 at 21:04
darklord0530
613
asked Nov 20 at 21:04
darklord0530
613
asked Nov 20 at 21:04
darklord0530
613
asked Nov 20 at 21:04
darklord0530
613
asked Nov 20 at 21:04
darklord0530
613
613
add a comment |
add a comment |
2 Answers
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You need to switch the two limits. In a way it is similar to saying if the solid is bounded by $x=2, x=0, ...$ You would not set up the integral as $int_2^0...$
answered Nov 20 at 21:08
Andrei
10.7k21025
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With reference to the following sketch
the set up should be
$$int_0^2 dx int_0^4 dz int_z^{z+1}dy$$
answered Nov 20 at 21:10
gimusi
92.7k94495
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You need to switch the two limits. In a way it is similar to saying if the solid is bounded by $x=2, x=0, ...$ You would not set up the integral as $int_2^0...$
answered Nov 20 at 21:08
Andrei
10.7k21025
add a comment |
up vote
1
down vote
accepted
You need to switch the two limits. In a way it is similar to saying if the solid is bounded by $x=2, x=0, ...$ You would not set up the integral as $int_2^0...$
answered Nov 20 at 21:08
Andrei
10.7k21025
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You need to switch the two limits. In a way it is similar to saying if the solid is bounded by $x=2, x=0, ...$ You would not set up the integral as $int_2^0...$
answered Nov 20 at 21:08
Andrei
10.7k21025
You need to switch the two limits. In a way it is similar to saying if the solid is bounded by $x=2, x=0, ...$ You would not set up the integral as $int_2^0...$
answered Nov 20 at 21:08
Andrei
10.7k21025
answered Nov 20 at 21:08
Andrei
10.7k21025
answered Nov 20 at 21:08
Andrei
10.7k21025
answered Nov 20 at 21:08
Andrei
10.7k21025
10.7k21025
add a comment |
add a comment |
up vote
1
down vote
With reference to the following sketch
the set up should be
$$int_0^2 dx int_0^4 dz int_z^{z+1}dy$$
answered Nov 20 at 21:10
gimusi
92.7k94495
add a comment |
up vote
1
down vote
With reference to the following sketch
the set up should be
$$int_0^2 dx int_0^4 dz int_z^{z+1}dy$$
answered Nov 20 at 21:10
gimusi
92.7k94495
add a comment |
up vote
1
down vote
up vote
1
down vote
With reference to the following sketch
the set up should be
$$int_0^2 dx int_0^4 dz int_z^{z+1}dy$$
answered Nov 20 at 21:10
gimusi
92.7k94495
With reference to the following sketch
the set up should be
$$int_0^2 dx int_0^4 dz int_z^{z+1}dy$$
answered Nov 20 at 21:10
gimusi
92.7k94495
edited Nov 20 at 21:14
answered Nov 20 at 21:10
gimusi
92.7k94495
answered Nov 20 at 21:10
gimusi
92.7k94495
answered Nov 20 at 21:10
gimusi
92.7k94495
92.7k94495
add a comment |
add a comment |
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