Limits for triple integral parabolic cylinder
Determine the volume bounded by the parabolic cylinder $z=x^2$ and the planes $y=0$ and $y+z=4$.
My work. I am not sure if I have the correct limits for this question. I used
$x = -sqrt{z},dots, sqrt{z}$,
$y= 0,dots, 4-z$,
$z=0,dots,4$.
It seems too easy, should I be using polars?
calculus multivariable-calculus
edited Dec 8 at 11:04
Robert Z
93.2k1061132
asked Dec 8 at 10:28
Pumpkinpeach
628
|
show 3 more comments
Determine the volume bounded by the parabolic cylinder $z=x^2$ and the planes $y=0$ and $y+z=4$.
My work. I am not sure if I have the correct limits for this question. I used
$x = -sqrt{z},dots, sqrt{z}$,
$y= 0,dots, 4-z$,
$z=0,dots,4$.
It seems too easy, should I be using polars?
calculus multivariable-calculus
edited Dec 8 at 11:04
Robert Z
93.2k1061132
asked Dec 8 at 10:28
Pumpkinpeach
628
Please do not ask questions using pictures of text, since otherwise the question is difficult to search for and some users cannot see the pictures on some devices.
– Shaun
Dec 8 at 10:38
Fair question! (+1)
– Robert Z
Dec 8 at 10:45
If it easy good for you! We don't really need polar in that case since for $z$ fixed the domain is rectangular.
– gimusi
Dec 8 at 10:52
I mean its a lot of marks and I felt I hadn't done enough work for it.
– Pumpkinpeach
Dec 8 at 10:54
1
@Pumpkinpeach To avoid downvotes, let us know your progress. BTW please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Dec 8 at 11:30
|
show 3 more comments
Determine the volume bounded by the parabolic cylinder $z=x^2$ and the planes $y=0$ and $y+z=4$.
My work. I am not sure if I have the correct limits for this question. I used
$x = -sqrt{z},dots, sqrt{z}$,
$y= 0,dots, 4-z$,
$z=0,dots,4$.
It seems too easy, should I be using polars?
calculus multivariable-calculus
edited Dec 8 at 11:04
Robert Z
93.2k1061132
asked Dec 8 at 10:28
Pumpkinpeach
628
Determine the volume bounded by the parabolic cylinder $z=x^2$ and the planes $y=0$ and $y+z=4$.
My work. I am not sure if I have the correct limits for this question. I used
$x = -sqrt{z},dots, sqrt{z}$,
$y= 0,dots, 4-z$,
$z=0,dots,4$.
It seems too easy, should I be using polars?
calculus multivariable-calculus
calculus multivariable-calculus
edited Dec 8 at 11:04
Robert Z
93.2k1061132
asked Dec 8 at 10:28
Pumpkinpeach
628
edited Dec 8 at 11:04
Robert Z
93.2k1061132
asked Dec 8 at 10:28
Pumpkinpeach
628
edited Dec 8 at 11:04
Robert Z
93.2k1061132
edited Dec 8 at 11:04
Robert Z
93.2k1061132
edited Dec 8 at 11:04
Robert Z
93.2k1061132
93.2k1061132
asked Dec 8 at 10:28
Pumpkinpeach
628
asked Dec 8 at 10:28
Pumpkinpeach
628
asked Dec 8 at 10:28
Pumpkinpeach
628
628
Please do not ask questions using pictures of text, since otherwise the question is difficult to search for and some users cannot see the pictures on some devices.
– Shaun
Dec 8 at 10:38
Fair question! (+1)
– Robert Z
Dec 8 at 10:45
If it easy good for you! We don't really need polar in that case since for $z$ fixed the domain is rectangular.
– gimusi
Dec 8 at 10:52
I mean its a lot of marks and I felt I hadn't done enough work for it.
– Pumpkinpeach
Dec 8 at 10:54
1
@Pumpkinpeach To avoid downvotes, let us know your progress. BTW please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Dec 8 at 11:30
|
show 3 more comments
Please do not ask questions using pictures of text, since otherwise the question is difficult to search for and some users cannot see the pictures on some devices.
– Shaun
Dec 8 at 10:38
Fair question! (+1)
– Robert Z
Dec 8 at 10:45
If it easy good for you! We don't really need polar in that case since for $z$ fixed the domain is rectangular.
– gimusi
Dec 8 at 10:52
I mean its a lot of marks and I felt I hadn't done enough work for it.
– Pumpkinpeach
Dec 8 at 10:54
1
@Pumpkinpeach To avoid downvotes, let us know your progress. BTW please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Dec 8 at 11:30
Please do not ask questions using pictures of text, since otherwise the question is difficult to search for and some users cannot see the pictures on some devices.
– Shaun
Dec 8 at 10:38
Please do not ask questions using pictures of text, since otherwise the question is difficult to search for and some users cannot see the pictures on some devices.
– Shaun
Dec 8 at 10:38
Fair question! (+1)
– Robert Z
Dec 8 at 10:45
Fair question! (+1)
– Robert Z
Dec 8 at 10:45
If it easy good for you! We don't really need polar in that case since for $z$ fixed the domain is rectangular.
– gimusi
Dec 8 at 10:52
If it easy good for you! We don't really need polar in that case since for $z$ fixed the domain is rectangular.
– gimusi
Dec 8 at 10:52
I mean its a lot of marks and I felt I hadn't done enough work for it.
– Pumpkinpeach
Dec 8 at 10:54
I mean its a lot of marks and I felt I hadn't done enough work for it.
– Pumpkinpeach
Dec 8 at 10:54
1
1
@Pumpkinpeach To avoid downvotes, let us know your progress. BTW please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Dec 8 at 11:30
@Pumpkinpeach To avoid downvotes, let us know your progress. BTW please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Dec 8 at 11:30
|
show 3 more comments
2 Answers
2
active
oldest
votes
Yes, the limits are correct, the volume is given by the following iterated integral
$$int_{z=0}^{4}left(int_{x=-sqrt{z}}^{sqrt{z}}left(int_{y=0}^{4-z} dyright) dxright) dz.$$
Can you take it from here? Cartesian coordinates seems to be fine here.
answered Dec 8 at 10:37
Robert Z
93.2k1061132
add a comment |
To check the set up we need to make some sketches of the domain as for example in $z-x$ and $z-y$ planes
and also important in the $x-y$ plane for $z$ fixed that is a rectangular domain
from here your set up seems correct.
answered Dec 8 at 10:47
gimusi
1
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
Yes, the limits are correct, the volume is given by the following iterated integral
$$int_{z=0}^{4}left(int_{x=-sqrt{z}}^{sqrt{z}}left(int_{y=0}^{4-z} dyright) dxright) dz.$$
Can you take it from here? Cartesian coordinates seems to be fine here.
answered Dec 8 at 10:37
Robert Z
93.2k1061132
add a comment |
Yes, the limits are correct, the volume is given by the following iterated integral
$$int_{z=0}^{4}left(int_{x=-sqrt{z}}^{sqrt{z}}left(int_{y=0}^{4-z} dyright) dxright) dz.$$
Can you take it from here? Cartesian coordinates seems to be fine here.
answered Dec 8 at 10:37
Robert Z
93.2k1061132
add a comment |
Yes, the limits are correct, the volume is given by the following iterated integral
$$int_{z=0}^{4}left(int_{x=-sqrt{z}}^{sqrt{z}}left(int_{y=0}^{4-z} dyright) dxright) dz.$$
Can you take it from here? Cartesian coordinates seems to be fine here.
answered Dec 8 at 10:37
Robert Z
93.2k1061132
Yes, the limits are correct, the volume is given by the following iterated integral
$$int_{z=0}^{4}left(int_{x=-sqrt{z}}^{sqrt{z}}left(int_{y=0}^{4-z} dyright) dxright) dz.$$
Can you take it from here? Cartesian coordinates seems to be fine here.
answered Dec 8 at 10:37
Robert Z
93.2k1061132
edited Dec 8 at 11:03
answered Dec 8 at 10:37
Robert Z
93.2k1061132
answered Dec 8 at 10:37
Robert Z
93.2k1061132
answered Dec 8 at 10:37
Robert Z
93.2k1061132
93.2k1061132
add a comment |
add a comment |
To check the set up we need to make some sketches of the domain as for example in $z-x$ and $z-y$ planes
and also important in the $x-y$ plane for $z$ fixed that is a rectangular domain
from here your set up seems correct.
answered Dec 8 at 10:47
gimusi
1
add a comment |
To check the set up we need to make some sketches of the domain as for example in $z-x$ and $z-y$ planes
and also important in the $x-y$ plane for $z$ fixed that is a rectangular domain
from here your set up seems correct.
answered Dec 8 at 10:47
gimusi
1
add a comment |
To check the set up we need to make some sketches of the domain as for example in $z-x$ and $z-y$ planes
and also important in the $x-y$ plane for $z$ fixed that is a rectangular domain
from here your set up seems correct.
answered Dec 8 at 10:47
gimusi
1
To check the set up we need to make some sketches of the domain as for example in $z-x$ and $z-y$ planes
and also important in the $x-y$ plane for $z$ fixed that is a rectangular domain
from here your set up seems correct.
answered Dec 8 at 10:47
gimusi
1
edited Dec 8 at 11:14
answered Dec 8 at 10:47
gimusi
1
answered Dec 8 at 10:47
gimusi
1
answered Dec 8 at 10:47
gimusi
1
1
add a comment |
add a comment |
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Please do not ask questions using pictures of text, since otherwise the question is difficult to search for and some users cannot see the pictures on some devices.
– Shaun
Dec 8 at 10:38
Fair question! (+1)
– Robert Z
Dec 8 at 10:45
If it easy good for you! We don't really need polar in that case since for $z$ fixed the domain is rectangular.
– gimusi
Dec 8 at 10:52
I mean its a lot of marks and I felt I hadn't done enough work for it.
– Pumpkinpeach
Dec 8 at 10:54
1
@Pumpkinpeach To avoid downvotes, let us know your progress. BTW please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Dec 8 at 11:30