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How to determine upper and lower limits of theta for cylindrical volume?
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Give the limits of integration for evaluating the integral $iiint_Rf(r,theta,z),dz,r,dr,dtheta$ as an iterated integral over the region that is bounded below by the plane $z=0$, on the side by the cylinder $r=9costheta$, and on top by the paraboloid $z=3r^2$.
$0le zle2r^2$; $0le rle9costheta$; what about $theta$?
What I don't understand is how to get the last one. I would think the lower limit of $ theta$ would be $0$, and its upper limit would be $2À$, considering both components of the 3D objects have the full range of rotation for $θ$. Apparently the answer is ACTUALLY that $-pi/2 ≤ theta ≤ pi /2$. I don't understand how to arrive at that answer.
definite-integrals volume cylindrical-coordinates
edited Jul 16 at 1:40
Parcly Taxel
33.6k136588
asked Jul 16 at 0:57
PCRevolt
326
add a comment |Â
up vote
-1
down vote
favorite
Give the limits of integration for evaluating the integral $iiint_Rf(r,theta,z),dz,r,dr,dtheta$ as an iterated integral over the region that is bounded below by the plane $z=0$, on the side by the cylinder $r=9costheta$, and on top by the paraboloid $z=3r^2$.
$0le zle2r^2$; $0le rle9costheta$; what about $theta$?
What I don't understand is how to get the last one. I would think the lower limit of $ theta$ would be $0$, and its upper limit would be $2À$, considering both components of the 3D objects have the full range of rotation for $θ$. Apparently the answer is ACTUALLY that $-pi/2 ≤ theta ≤ pi /2$. I don't understand how to arrive at that answer.
definite-integrals volume cylindrical-coordinates
edited Jul 16 at 1:40
Parcly Taxel
33.6k136588
asked Jul 16 at 0:57
PCRevolt
326
How can you have $0le r le 9costheta$ when $costheta < 0?$
– saulspatz
Jul 16 at 1:44
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Give the limits of integration for evaluating the integral $iiint_Rf(r,theta,z),dz,r,dr,dtheta$ as an iterated integral over the region that is bounded below by the plane $z=0$, on the side by the cylinder $r=9costheta$, and on top by the paraboloid $z=3r^2$.
$0le zle2r^2$; $0le rle9costheta$; what about $theta$?
What I don't understand is how to get the last one. I would think the lower limit of $ theta$ would be $0$, and its upper limit would be $2À$, considering both components of the 3D objects have the full range of rotation for $θ$. Apparently the answer is ACTUALLY that $-pi/2 ≤ theta ≤ pi /2$. I don't understand how to arrive at that answer.
definite-integrals volume cylindrical-coordinates
edited Jul 16 at 1:40
Parcly Taxel
33.6k136588
asked Jul 16 at 0:57
PCRevolt
326
Give the limits of integration for evaluating the integral $iiint_Rf(r,theta,z),dz,r,dr,dtheta$ as an iterated integral over the region that is bounded below by the plane $z=0$, on the side by the cylinder $r=9costheta$, and on top by the paraboloid $z=3r^2$.
$0le zle2r^2$; $0le rle9costheta$; what about $theta$?
What I don't understand is how to get the last one. I would think the lower limit of $ theta$ would be $0$, and its upper limit would be $2À$, considering both components of the 3D objects have the full range of rotation for $θ$. Apparently the answer is ACTUALLY that $-pi/2 ≤ theta ≤ pi /2$. I don't understand how to arrive at that answer.
definite-integrals volume cylindrical-coordinates
edited Jul 16 at 1:40
Parcly Taxel
33.6k136588
asked Jul 16 at 0:57
PCRevolt
326
edited Jul 16 at 1:40
Parcly Taxel
33.6k136588
edited Jul 16 at 1:40
Parcly Taxel
33.6k136588
edited Jul 16 at 1:40
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 16 at 0:57
PCRevolt
326
asked Jul 16 at 0:57
PCRevolt
326
asked Jul 16 at 0:57
PCRevolt
326
326
How can you have $0le r le 9costheta$ when $costheta < 0?$
– saulspatz
Jul 16 at 1:44
add a comment |Â
How can you have $0le r le 9costheta$ when $costheta < 0?$
– saulspatz
Jul 16 at 1:44
How can you have $0le r le 9costheta$ when $costheta < 0?$
– saulspatz
Jul 16 at 1:44
How can you have $0le r le 9costheta$ when $costheta < 0?$
– saulspatz
Jul 16 at 1:44
add a comment |Â
1 Answer
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The plot of your region is:
I've drawn the cylinder as a spiral in order not to cover the inside. Let's just look at the equation of your cylinder:
$$r = 9 cos(theta) $$
$$ r^2 = 9rcos(theta)$$
Since $r^2 = x^2 + y^2$ and $x = r cos(theta)$ we have:
$$x^2 + y^2 = 9x$$
$$(x - 4.5)^2 + y^2 = 4.5^2 = r^2$$
From this we can conclude that the angle $theta$ which lies on the $xy$ plane is bounded by $-fracpi2$ and $fracpi2$
As you can see the argument of $x_1$ is $pi /2$ and the argument of $x_2$ is $-pi /2$ (well, almost).
answered Jul 16 at 2:36
user1949350
977
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The plot of your region is:
I've drawn the cylinder as a spiral in order not to cover the inside. Let's just look at the equation of your cylinder:
$$r = 9 cos(theta) $$
$$ r^2 = 9rcos(theta)$$
Since $r^2 = x^2 + y^2$ and $x = r cos(theta)$ we have:
$$x^2 + y^2 = 9x$$
$$(x - 4.5)^2 + y^2 = 4.5^2 = r^2$$
From this we can conclude that the angle $theta$ which lies on the $xy$ plane is bounded by $-fracpi2$ and $fracpi2$
As you can see the argument of $x_1$ is $pi /2$ and the argument of $x_2$ is $-pi /2$ (well, almost).
answered Jul 16 at 2:36
user1949350
977
add a comment |Â
up vote
0
down vote
accepted
The plot of your region is:
I've drawn the cylinder as a spiral in order not to cover the inside. Let's just look at the equation of your cylinder:
$$r = 9 cos(theta) $$
$$ r^2 = 9rcos(theta)$$
Since $r^2 = x^2 + y^2$ and $x = r cos(theta)$ we have:
$$x^2 + y^2 = 9x$$
$$(x - 4.5)^2 + y^2 = 4.5^2 = r^2$$
From this we can conclude that the angle $theta$ which lies on the $xy$ plane is bounded by $-fracpi2$ and $fracpi2$
As you can see the argument of $x_1$ is $pi /2$ and the argument of $x_2$ is $-pi /2$ (well, almost).
answered Jul 16 at 2:36
user1949350
977
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The plot of your region is:
I've drawn the cylinder as a spiral in order not to cover the inside. Let's just look at the equation of your cylinder:
$$r = 9 cos(theta) $$
$$ r^2 = 9rcos(theta)$$
Since $r^2 = x^2 + y^2$ and $x = r cos(theta)$ we have:
$$x^2 + y^2 = 9x$$
$$(x - 4.5)^2 + y^2 = 4.5^2 = r^2$$
From this we can conclude that the angle $theta$ which lies on the $xy$ plane is bounded by $-fracpi2$ and $fracpi2$
As you can see the argument of $x_1$ is $pi /2$ and the argument of $x_2$ is $-pi /2$ (well, almost).
answered Jul 16 at 2:36
user1949350
977
The plot of your region is:
I've drawn the cylinder as a spiral in order not to cover the inside. Let's just look at the equation of your cylinder:
$$r = 9 cos(theta) $$
$$ r^2 = 9rcos(theta)$$
Since $r^2 = x^2 + y^2$ and $x = r cos(theta)$ we have:
$$x^2 + y^2 = 9x$$
$$(x - 4.5)^2 + y^2 = 4.5^2 = r^2$$
From this we can conclude that the angle $theta$ which lies on the $xy$ plane is bounded by $-fracpi2$ and $fracpi2$
As you can see the argument of $x_1$ is $pi /2$ and the argument of $x_2$ is $-pi /2$ (well, almost).
answered Jul 16 at 2:36
user1949350
977
edited Jul 16 at 2:47
answered Jul 16 at 2:36
user1949350
977
answered Jul 16 at 2:36
user1949350
977
answered Jul 16 at 2:36
user1949350
977
977
add a comment |Â
add a comment |Â
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How can you have $0le r le 9costheta$ when $costheta < 0?$
– saulspatz
Jul 16 at 1:44