Mixing
Calculation of triple integral
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Consider the 3D region R bounded by $x+y+z = 1$, $y =0$, $z=0$. Evaluate $∫∫∫(x^2 + y^2 + z^2 ) dx dy dz$ over the region $R$.
I tried to solve this question by taking limits of integration as
$$x: 0 to 1$$
$$y: 0 to 1-x$$
$$z: 0 to 1-x$$
I got the final answer as 29/180. But I am not sure whether this answer is correct or not. Please help me.
calculus
edited Jul 26 at 14:16
Nosrati
19.2k41544
asked Jul 26 at 14:12
Balaji
6617
add a comment |Â
up vote
0
down vote
favorite
Consider the 3D region R bounded by $x+y+z = 1$, $y =0$, $z=0$. Evaluate $∫∫∫(x^2 + y^2 + z^2 ) dx dy dz$ over the region $R$.
I tried to solve this question by taking limits of integration as
$$x: 0 to 1$$
$$y: 0 to 1-x$$
$$z: 0 to 1-x$$
I got the final answer as 29/180. But I am not sure whether this answer is correct or not. Please help me.
calculus
edited Jul 26 at 14:16
Nosrati
19.2k41544
asked Jul 26 at 14:12
Balaji
6617
2
Isn't also $x=0$ in your conditions?
– Nosrati
Jul 26 at 14:19
No, x=0 is not present
– Balaji
Jul 26 at 14:23
Without the plane $x=0$ the region $R$ is unbounded. In fact, I don't see a way of selecting which of the eight convex sets $R$ is supposed to be in that case? The integral diverges for each and every one of them. Please, check again.
– Jyrki Lahtonen
Jul 27 at 6:20
1
@Balaji Yes as noted by JL I’ve assumed that x=0 is a constraint otherwise the integral diverges. Check that point again, if you are not dealing with simple improper integrals it is probably a typo in your reference book or notes.
– gimusi
Jul 27 at 6:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the 3D region R bounded by $x+y+z = 1$, $y =0$, $z=0$. Evaluate $∫∫∫(x^2 + y^2 + z^2 ) dx dy dz$ over the region $R$.
I tried to solve this question by taking limits of integration as
$$x: 0 to 1$$
$$y: 0 to 1-x$$
$$z: 0 to 1-x$$
I got the final answer as 29/180. But I am not sure whether this answer is correct or not. Please help me.
calculus
edited Jul 26 at 14:16
Nosrati
19.2k41544
asked Jul 26 at 14:12
Balaji
6617
Consider the 3D region R bounded by $x+y+z = 1$, $y =0$, $z=0$. Evaluate $∫∫∫(x^2 + y^2 + z^2 ) dx dy dz$ over the region $R$.
I tried to solve this question by taking limits of integration as
$$x: 0 to 1$$
$$y: 0 to 1-x$$
$$z: 0 to 1-x$$
I got the final answer as 29/180. But I am not sure whether this answer is correct or not. Please help me.
calculus
edited Jul 26 at 14:16
Nosrati
19.2k41544
asked Jul 26 at 14:12
Balaji
6617
edited Jul 26 at 14:16
Nosrati
19.2k41544
edited Jul 26 at 14:16
Nosrati
19.2k41544
edited Jul 26 at 14:16
Nosrati
19.2k41544
19.2k41544
asked Jul 26 at 14:12
Balaji
6617
asked Jul 26 at 14:12
Balaji
6617
asked Jul 26 at 14:12
Balaji
6617
6617
2
Isn't also $x=0$ in your conditions?
– Nosrati
Jul 26 at 14:19
No, x=0 is not present
– Balaji
Jul 26 at 14:23
Without the plane $x=0$ the region $R$ is unbounded. In fact, I don't see a way of selecting which of the eight convex sets $R$ is supposed to be in that case? The integral diverges for each and every one of them. Please, check again.
– Jyrki Lahtonen
Jul 27 at 6:20
1
@Balaji Yes as noted by JL I’ve assumed that x=0 is a constraint otherwise the integral diverges. Check that point again, if you are not dealing with simple improper integrals it is probably a typo in your reference book or notes.
– gimusi
Jul 27 at 6:33
add a comment |Â
2
Isn't also $x=0$ in your conditions?
– Nosrati
Jul 26 at 14:19
No, x=0 is not present
– Balaji
Jul 26 at 14:23
Without the plane $x=0$ the region $R$ is unbounded. In fact, I don't see a way of selecting which of the eight convex sets $R$ is supposed to be in that case? The integral diverges for each and every one of them. Please, check again.
– Jyrki Lahtonen
Jul 27 at 6:20
1
@Balaji Yes as noted by JL I’ve assumed that x=0 is a constraint otherwise the integral diverges. Check that point again, if you are not dealing with simple improper integrals it is probably a typo in your reference book or notes.
– gimusi
Jul 27 at 6:33
2
2
Isn't also $x=0$ in your conditions?
– Nosrati
Jul 26 at 14:19
Isn't also $x=0$ in your conditions?
– Nosrati
Jul 26 at 14:19
No, x=0 is not present
– Balaji
Jul 26 at 14:23
No, x=0 is not present
– Balaji
Jul 26 at 14:23
Without the plane $x=0$ the region $R$ is unbounded. In fact, I don't see a way of selecting which of the eight convex sets $R$ is supposed to be in that case? The integral diverges for each and every one of them. Please, check again.
– Jyrki Lahtonen
Jul 27 at 6:20
Without the plane $x=0$ the region $R$ is unbounded. In fact, I don't see a way of selecting which of the eight convex sets $R$ is supposed to be in that case? The integral diverges for each and every one of them. Please, check again.
– Jyrki Lahtonen
Jul 27 at 6:20
1
1
@Balaji Yes as noted by JL I’ve assumed that x=0 is a constraint otherwise the integral diverges. Check that point again, if you are not dealing with simple improper integrals it is probably a typo in your reference book or notes.
– gimusi
Jul 27 at 6:33
@Balaji Yes as noted by JL I’ve assumed that x=0 is a constraint otherwise the integral diverges. Check that point again, if you are not dealing with simple improper integrals it is probably a typo in your reference book or notes.
– gimusi
Jul 27 at 6:33
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
-2
down vote
accepted
Assuming also $x=0$ as bound (otherwise the integral diverges), the correct set up should be
$$int_0^1 dxint_0^1-xdy int_0^1-x-y x^2 + y^2 + z^2 ,dz$$
and the final result is equal to 1/20, indeed we have
$$int_0^1-x-y x^2 + y^2 + z^2 ,dz=[x^2z+y^2z+z^3/3]_0^1-x-y=\=x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3$$
and
$$int_0^1-x x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3 dy=\=[x^2(1-x)-y^2x^2/2+(1-x)y^3/3-y^4/4-(1-x-y)^4/4]_0^1-x=\=x^2(1-x)-frac12(1-x)^4+frac13(1-x)^4-frac14(1-x)^4+frac14(1-x)^4=x^2(1-x)-frac16(1-x)^4$$
and
$$int_0^1 x^2(1-x)-frac 1 6(1-x)^4 dx=\=[x^3/3-x^4/4+(1-x)^5/30]_0^1=frac13-frac14-frac130=frac20-15-260=frac360=frac120$$
answered Jul 26 at 14:25
gimusi
65k73583
can you share your complete work as the calculation is very cumbersome. Thanks
– Balaji
Jul 26 at 15:00
@Balaji I've just added some detail.
– gimusi
Jul 26 at 15:29
Thanks a lot :)
– Balaji
Jul 26 at 16:51
You are welcome! Bye
– gimusi
Jul 26 at 16:52
1
This is just wrong. The point $(x,y,z)=(-2,1,1)$ satisfies the constraints $x+y+zle1$, $yge0$, $zge0$ but is left out from your calculation.
– Jyrki Lahtonen
Jul 27 at 6:22
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-2
down vote
accepted
Assuming also $x=0$ as bound (otherwise the integral diverges), the correct set up should be
$$int_0^1 dxint_0^1-xdy int_0^1-x-y x^2 + y^2 + z^2 ,dz$$
and the final result is equal to 1/20, indeed we have
$$int_0^1-x-y x^2 + y^2 + z^2 ,dz=[x^2z+y^2z+z^3/3]_0^1-x-y=\=x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3$$
and
$$int_0^1-x x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3 dy=\=[x^2(1-x)-y^2x^2/2+(1-x)y^3/3-y^4/4-(1-x-y)^4/4]_0^1-x=\=x^2(1-x)-frac12(1-x)^4+frac13(1-x)^4-frac14(1-x)^4+frac14(1-x)^4=x^2(1-x)-frac16(1-x)^4$$
and
$$int_0^1 x^2(1-x)-frac 1 6(1-x)^4 dx=\=[x^3/3-x^4/4+(1-x)^5/30]_0^1=frac13-frac14-frac130=frac20-15-260=frac360=frac120$$
answered Jul 26 at 14:25
gimusi
65k73583
can you share your complete work as the calculation is very cumbersome. Thanks
– Balaji
Jul 26 at 15:00
@Balaji I've just added some detail.
– gimusi
Jul 26 at 15:29
Thanks a lot :)
– Balaji
Jul 26 at 16:51
You are welcome! Bye
– gimusi
Jul 26 at 16:52
1
This is just wrong. The point $(x,y,z)=(-2,1,1)$ satisfies the constraints $x+y+zle1$, $yge0$, $zge0$ but is left out from your calculation.
– Jyrki Lahtonen
Jul 27 at 6:22
 |Â
show 2 more comments
up vote
-2
down vote
accepted
Assuming also $x=0$ as bound (otherwise the integral diverges), the correct set up should be
$$int_0^1 dxint_0^1-xdy int_0^1-x-y x^2 + y^2 + z^2 ,dz$$
and the final result is equal to 1/20, indeed we have
$$int_0^1-x-y x^2 + y^2 + z^2 ,dz=[x^2z+y^2z+z^3/3]_0^1-x-y=\=x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3$$
and
$$int_0^1-x x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3 dy=\=[x^2(1-x)-y^2x^2/2+(1-x)y^3/3-y^4/4-(1-x-y)^4/4]_0^1-x=\=x^2(1-x)-frac12(1-x)^4+frac13(1-x)^4-frac14(1-x)^4+frac14(1-x)^4=x^2(1-x)-frac16(1-x)^4$$
and
$$int_0^1 x^2(1-x)-frac 1 6(1-x)^4 dx=\=[x^3/3-x^4/4+(1-x)^5/30]_0^1=frac13-frac14-frac130=frac20-15-260=frac360=frac120$$
answered Jul 26 at 14:25
gimusi
65k73583
can you share your complete work as the calculation is very cumbersome. Thanks
– Balaji
Jul 26 at 15:00
@Balaji I've just added some detail.
– gimusi
Jul 26 at 15:29
Thanks a lot :)
– Balaji
Jul 26 at 16:51
You are welcome! Bye
– gimusi
Jul 26 at 16:52
1
This is just wrong. The point $(x,y,z)=(-2,1,1)$ satisfies the constraints $x+y+zle1$, $yge0$, $zge0$ but is left out from your calculation.
– Jyrki Lahtonen
Jul 27 at 6:22
 |Â
show 2 more comments
up vote
-2
down vote
accepted
up vote
-2
down vote
accepted
Assuming also $x=0$ as bound (otherwise the integral diverges), the correct set up should be
$$int_0^1 dxint_0^1-xdy int_0^1-x-y x^2 + y^2 + z^2 ,dz$$
and the final result is equal to 1/20, indeed we have
$$int_0^1-x-y x^2 + y^2 + z^2 ,dz=[x^2z+y^2z+z^3/3]_0^1-x-y=\=x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3$$
and
$$int_0^1-x x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3 dy=\=[x^2(1-x)-y^2x^2/2+(1-x)y^3/3-y^4/4-(1-x-y)^4/4]_0^1-x=\=x^2(1-x)-frac12(1-x)^4+frac13(1-x)^4-frac14(1-x)^4+frac14(1-x)^4=x^2(1-x)-frac16(1-x)^4$$
and
$$int_0^1 x^2(1-x)-frac 1 6(1-x)^4 dx=\=[x^3/3-x^4/4+(1-x)^5/30]_0^1=frac13-frac14-frac130=frac20-15-260=frac360=frac120$$
answered Jul 26 at 14:25
gimusi
65k73583
Assuming also $x=0$ as bound (otherwise the integral diverges), the correct set up should be
$$int_0^1 dxint_0^1-xdy int_0^1-x-y x^2 + y^2 + z^2 ,dz$$
and the final result is equal to 1/20, indeed we have
$$int_0^1-x-y x^2 + y^2 + z^2 ,dz=[x^2z+y^2z+z^3/3]_0^1-x-y=\=x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3$$
and
$$int_0^1-x x^2(1-x-y)+y^2(1-x-y)+frac13(1-x-y)^3 dy=\=[x^2(1-x)-y^2x^2/2+(1-x)y^3/3-y^4/4-(1-x-y)^4/4]_0^1-x=\=x^2(1-x)-frac12(1-x)^4+frac13(1-x)^4-frac14(1-x)^4+frac14(1-x)^4=x^2(1-x)-frac16(1-x)^4$$
and
$$int_0^1 x^2(1-x)-frac 1 6(1-x)^4 dx=\=[x^3/3-x^4/4+(1-x)^5/30]_0^1=frac13-frac14-frac130=frac20-15-260=frac360=frac120$$
answered Jul 26 at 14:25
gimusi
65k73583
edited Jul 27 at 17:57
answered Jul 26 at 14:25
gimusi
65k73583
answered Jul 26 at 14:25
gimusi
65k73583
answered Jul 26 at 14:25
gimusi
65k73583
65k73583
can you share your complete work as the calculation is very cumbersome. Thanks
– Balaji
Jul 26 at 15:00
@Balaji I've just added some detail.
– gimusi
Jul 26 at 15:29
Thanks a lot :)
– Balaji
Jul 26 at 16:51
You are welcome! Bye
– gimusi
Jul 26 at 16:52
1
This is just wrong. The point $(x,y,z)=(-2,1,1)$ satisfies the constraints $x+y+zle1$, $yge0$, $zge0$ but is left out from your calculation.
– Jyrki Lahtonen
Jul 27 at 6:22
 |Â
show 2 more comments
can you share your complete work as the calculation is very cumbersome. Thanks
– Balaji
Jul 26 at 15:00
@Balaji I've just added some detail.
– gimusi
Jul 26 at 15:29
Thanks a lot :)
– Balaji
Jul 26 at 16:51
You are welcome! Bye
– gimusi
Jul 26 at 16:52
1
This is just wrong. The point $(x,y,z)=(-2,1,1)$ satisfies the constraints $x+y+zle1$, $yge0$, $zge0$ but is left out from your calculation.
– Jyrki Lahtonen
Jul 27 at 6:22
can you share your complete work as the calculation is very cumbersome. Thanks
– Balaji
Jul 26 at 15:00
can you share your complete work as the calculation is very cumbersome. Thanks
– Balaji
Jul 26 at 15:00
@Balaji I've just added some detail.
– gimusi
Jul 26 at 15:29
@Balaji I've just added some detail.
– gimusi
Jul 26 at 15:29
Thanks a lot :)
– Balaji
Jul 26 at 16:51
Thanks a lot :)
– Balaji
Jul 26 at 16:51
You are welcome! Bye
– gimusi
Jul 26 at 16:52
You are welcome! Bye
– gimusi
Jul 26 at 16:52
1
1
This is just wrong. The point $(x,y,z)=(-2,1,1)$ satisfies the constraints $x+y+zle1$, $yge0$, $zge0$ but is left out from your calculation.
– Jyrki Lahtonen
Jul 27 at 6:22
This is just wrong. The point $(x,y,z)=(-2,1,1)$ satisfies the constraints $x+y+zle1$, $yge0$, $zge0$ but is left out from your calculation.
– Jyrki Lahtonen
Jul 27 at 6:22
 |Â
show 2 more comments
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2
Isn't also $x=0$ in your conditions?
– Nosrati
Jul 26 at 14:19
No, x=0 is not present
– Balaji
Jul 26 at 14:23
Without the plane $x=0$ the region $R$ is unbounded. In fact, I don't see a way of selecting which of the eight convex sets $R$ is supposed to be in that case? The integral diverges for each and every one of them. Please, check again.
– Jyrki Lahtonen
Jul 27 at 6:20
1
@Balaji Yes as noted by JL I’ve assumed that x=0 is a constraint otherwise the integral diverges. Check that point again, if you are not dealing with simple improper integrals it is probably a typo in your reference book or notes.
– gimusi
Jul 27 at 6:33