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Volume of the section surrounded by functions…
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Find the Volume of the section surrounded by functions
$z=2x^2+2y^2,quad x^2+y^2=2x,quad z=0$.
When converted these to $z=2x^2+2y^2, (x-1)^2+y^2=1, z=0$
,
I thought it may get easier if I convert Cartesian to cylindrical coordinates.
The results are $f_1: z=2r^2$ (?), $f_2: r=2costheta$ and I tried this and failed.
$$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
What makes me confused is whether setting $z=2costheta$ is possible when $r=2costheta$.
Where did I do wrong?
integration
asked Jul 21 at 9:06
NK Yu
1847
add a comment |Â
up vote
-1
down vote
favorite
Find the Volume of the section surrounded by functions
$z=2x^2+2y^2,quad x^2+y^2=2x,quad z=0$.
When converted these to $z=2x^2+2y^2, (x-1)^2+y^2=1, z=0$
,
I thought it may get easier if I convert Cartesian to cylindrical coordinates.
The results are $f_1: z=2r^2$ (?), $f_2: r=2costheta$ and I tried this and failed.
$$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
What makes me confused is whether setting $z=2costheta$ is possible when $r=2costheta$.
Where did I do wrong?
integration
asked Jul 21 at 9:06
NK Yu
1847
It's not clear to me, are you trying to evaluate the integral $$iiint_D2x^2+2y^2dxdydz$$ where $D$ is the domain defined by $$D=left(x,y,z);;texts.t.;; x^2+y^2=2x;;textand;;z=0right$$
– Davide Morgante
Jul 21 at 9:10
yes. I think so.
– NK Yu
Jul 22 at 10:27
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Find the Volume of the section surrounded by functions
$z=2x^2+2y^2,quad x^2+y^2=2x,quad z=0$.
When converted these to $z=2x^2+2y^2, (x-1)^2+y^2=1, z=0$
,
I thought it may get easier if I convert Cartesian to cylindrical coordinates.
The results are $f_1: z=2r^2$ (?), $f_2: r=2costheta$ and I tried this and failed.
$$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
What makes me confused is whether setting $z=2costheta$ is possible when $r=2costheta$.
Where did I do wrong?
integration
asked Jul 21 at 9:06
NK Yu
1847
Find the Volume of the section surrounded by functions
$z=2x^2+2y^2,quad x^2+y^2=2x,quad z=0$.
When converted these to $z=2x^2+2y^2, (x-1)^2+y^2=1, z=0$
,
I thought it may get easier if I convert Cartesian to cylindrical coordinates.
The results are $f_1: z=2r^2$ (?), $f_2: r=2costheta$ and I tried this and failed.
$$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
What makes me confused is whether setting $z=2costheta$ is possible when $r=2costheta$.
Where did I do wrong?
integration
asked Jul 21 at 9:06
NK Yu
1847
edited Jul 28 at 0:08
asked Jul 21 at 9:06
NK Yu
1847
asked Jul 21 at 9:06
NK Yu
1847
asked Jul 21 at 9:06
NK Yu
1847
1847
It's not clear to me, are you trying to evaluate the integral $$iiint_D2x^2+2y^2dxdydz$$ where $D$ is the domain defined by $$D=left(x,y,z);;texts.t.;; x^2+y^2=2x;;textand;;z=0right$$
– Davide Morgante
Jul 21 at 9:10
yes. I think so.
– NK Yu
Jul 22 at 10:27
add a comment |Â
It's not clear to me, are you trying to evaluate the integral $$iiint_D2x^2+2y^2dxdydz$$ where $D$ is the domain defined by $$D=left(x,y,z);;texts.t.;; x^2+y^2=2x;;textand;;z=0right$$
– Davide Morgante
Jul 21 at 9:10
yes. I think so.
– NK Yu
Jul 22 at 10:27
It's not clear to me, are you trying to evaluate the integral $$iiint_D2x^2+2y^2dxdydz$$ where $D$ is the domain defined by $$D=left(x,y,z);;texts.t.;; x^2+y^2=2x;;textand;;z=0right$$
– Davide Morgante
Jul 21 at 9:10
It's not clear to me, are you trying to evaluate the integral $$iiint_D2x^2+2y^2dxdydz$$ where $D$ is the domain defined by $$D=left(x,y,z);;texts.t.;; x^2+y^2=2x;;textand;;z=0right$$
– Davide Morgante
Jul 21 at 9:10
yes. I think so.
– NK Yu
Jul 22 at 10:27
yes. I think so.
– NK Yu
Jul 22 at 10:27
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your limits of $ theta $ $$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
are questionable.
Also the function $2r^2$ in the integrand is not necessary because you just want to find the volume.
Otherwise your integral seems to be fine.
answered Jul 21 at 10:37
Mohammad Riazi-Kermani
27.5k41852
how can I get the volume under z=2x^2+2y^2 if not for 2r^2?
– NK Yu
Jul 22 at 10:26
For the volume your function is $1$ and $2r^2$ appears in the upper limit.
– Mohammad Riazi-Kermani
Jul 22 at 10:29
I gets 3pi. but you said limits of θ are questionable. how so?
– NK Yu
Jul 28 at 0:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your limits of $ theta $ $$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
are questionable.
Also the function $2r^2$ in the integrand is not necessary because you just want to find the volume.
Otherwise your integral seems to be fine.
answered Jul 21 at 10:37
Mohammad Riazi-Kermani
27.5k41852
how can I get the volume under z=2x^2+2y^2 if not for 2r^2?
– NK Yu
Jul 22 at 10:26
For the volume your function is $1$ and $2r^2$ appears in the upper limit.
– Mohammad Riazi-Kermani
Jul 22 at 10:29
I gets 3pi. but you said limits of θ are questionable. how so?
– NK Yu
Jul 28 at 0:34
add a comment |Â
up vote
1
down vote
accepted
Your limits of $ theta $ $$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
are questionable.
Also the function $2r^2$ in the integrand is not necessary because you just want to find the volume.
Otherwise your integral seems to be fine.
answered Jul 21 at 10:37
Mohammad Riazi-Kermani
27.5k41852
how can I get the volume under z=2x^2+2y^2 if not for 2r^2?
– NK Yu
Jul 22 at 10:26
For the volume your function is $1$ and $2r^2$ appears in the upper limit.
– Mohammad Riazi-Kermani
Jul 22 at 10:29
I gets 3pi. but you said limits of θ are questionable. how so?
– NK Yu
Jul 28 at 0:34
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your limits of $ theta $ $$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
are questionable.
Also the function $2r^2$ in the integrand is not necessary because you just want to find the volume.
Otherwise your integral seems to be fine.
answered Jul 21 at 10:37
Mohammad Riazi-Kermani
27.5k41852
Your limits of $ theta $ $$int_0^2piint_0^2costhetaint_0^2r^22r^2r,dz,dr,dtheta$$
are questionable.
Also the function $2r^2$ in the integrand is not necessary because you just want to find the volume.
Otherwise your integral seems to be fine.
answered Jul 21 at 10:37
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 10:37
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 10:37
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 10:37
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
how can I get the volume under z=2x^2+2y^2 if not for 2r^2?
– NK Yu
Jul 22 at 10:26
For the volume your function is $1$ and $2r^2$ appears in the upper limit.
– Mohammad Riazi-Kermani
Jul 22 at 10:29
I gets 3pi. but you said limits of θ are questionable. how so?
– NK Yu
Jul 28 at 0:34
add a comment |Â
how can I get the volume under z=2x^2+2y^2 if not for 2r^2?
– NK Yu
Jul 22 at 10:26
For the volume your function is $1$ and $2r^2$ appears in the upper limit.
– Mohammad Riazi-Kermani
Jul 22 at 10:29
I gets 3pi. but you said limits of θ are questionable. how so?
– NK Yu
Jul 28 at 0:34
how can I get the volume under z=2x^2+2y^2 if not for 2r^2?
– NK Yu
Jul 22 at 10:26
how can I get the volume under z=2x^2+2y^2 if not for 2r^2?
– NK Yu
Jul 22 at 10:26
For the volume your function is $1$ and $2r^2$ appears in the upper limit.
– Mohammad Riazi-Kermani
Jul 22 at 10:29
For the volume your function is $1$ and $2r^2$ appears in the upper limit.
– Mohammad Riazi-Kermani
Jul 22 at 10:29
I gets 3pi. but you said limits of θ are questionable. how so?
– NK Yu
Jul 28 at 0:34
I gets 3pi. but you said limits of θ are questionable. how so?
– NK Yu
Jul 28 at 0:34
add a comment |Â
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It's not clear to me, are you trying to evaluate the integral $$iiint_D2x^2+2y^2dxdydz$$ where $D$ is the domain defined by $$D=left(x,y,z);;texts.t.;; x^2+y^2=2x;;textand;;z=0right$$
– Davide Morgante
Jul 21 at 9:10
yes. I think so.
– NK Yu
Jul 22 at 10:27