Evaluate this triple integral (volume)
$begingroup$
I need to calculate this volume of
$$D={x^2+y^2-2yle 0,0le zle 10-3sqrt{x^2+y^2} }.$$
My attempt. So the first one is a shifted cylinder $x^2+(y-1)^2le 1$ , and the second one is an upside-down cone with vertex $z=10$.
By using cylindrical coordinates I get:
$$int_{0}^{2pi}int_{0}^{sqrt{2sin{theta}}}int_{0}^{10-3r}rdzdrdtheta$$
Is this the correct way to approach a shifted region? $x^2+y^2=2y$ implies $r^2=2sin{theta}$.
I think I got the integral wrong.
calculus integration multivariable-calculus definite-integrals
edited Jan 16 at 8:03
Robert Z
101k1071144
asked Jan 16 at 7:43
NPLSNPLS
7812
$endgroup$
add a comment |
$begingroup$
I need to calculate this volume of
$$D={x^2+y^2-2yle 0,0le zle 10-3sqrt{x^2+y^2} }.$$
My attempt. So the first one is a shifted cylinder $x^2+(y-1)^2le 1$ , and the second one is an upside-down cone with vertex $z=10$.
By using cylindrical coordinates I get:
$$int_{0}^{2pi}int_{0}^{sqrt{2sin{theta}}}int_{0}^{10-3r}rdzdrdtheta$$
Is this the correct way to approach a shifted region? $x^2+y^2=2y$ implies $r^2=2sin{theta}$.
I think I got the integral wrong.
calculus integration multivariable-calculus definite-integrals
edited Jan 16 at 8:03
Robert Z
101k1071144
asked Jan 16 at 7:43
NPLSNPLS
7812
$endgroup$
add a comment |
$begingroup$
I need to calculate this volume of
$$D={x^2+y^2-2yle 0,0le zle 10-3sqrt{x^2+y^2} }.$$
My attempt. So the first one is a shifted cylinder $x^2+(y-1)^2le 1$ , and the second one is an upside-down cone with vertex $z=10$.
By using cylindrical coordinates I get:
$$int_{0}^{2pi}int_{0}^{sqrt{2sin{theta}}}int_{0}^{10-3r}rdzdrdtheta$$
Is this the correct way to approach a shifted region? $x^2+y^2=2y$ implies $r^2=2sin{theta}$.
I think I got the integral wrong.
calculus integration multivariable-calculus definite-integrals
edited Jan 16 at 8:03
Robert Z
101k1071144
asked Jan 16 at 7:43
NPLSNPLS
7812
$endgroup$
I need to calculate this volume of
$$D={x^2+y^2-2yle 0,0le zle 10-3sqrt{x^2+y^2} }.$$
My attempt. So the first one is a shifted cylinder $x^2+(y-1)^2le 1$ , and the second one is an upside-down cone with vertex $z=10$.
By using cylindrical coordinates I get:
$$int_{0}^{2pi}int_{0}^{sqrt{2sin{theta}}}int_{0}^{10-3r}rdzdrdtheta$$
Is this the correct way to approach a shifted region? $x^2+y^2=2y$ implies $r^2=2sin{theta}$.
I think I got the integral wrong.
calculus integration multivariable-calculus definite-integrals
calculus integration multivariable-calculus definite-integrals
edited Jan 16 at 8:03
Robert Z
101k1071144
asked Jan 16 at 7:43
NPLSNPLS
7812
edited Jan 16 at 8:03
Robert Z
101k1071144
asked Jan 16 at 7:43
NPLSNPLS
7812
edited Jan 16 at 8:03
Robert Z
101k1071144
edited Jan 16 at 8:03
Robert Z
101k1071144
edited Jan 16 at 8:03
Robert Z
101k1071144
101k1071144
asked Jan 16 at 7:43
NPLSNPLS
7812
asked Jan 16 at 7:43
NPLSNPLS
7812
asked Jan 16 at 7:43
NPLSNPLS
7812
7812
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You are almost correct. The angle $theta$ should be in the interval $[0,pi]$, so $0leq rleq 2sinthetageq 0$ (note that $r^2leq 2rsintheta$). Thus your triple integral should be
$$int_{0}^{pi}int_{0}^{2sin{theta}}int_{0}^{10-3r}rdzdrdtheta.$$
So what is the final result?
answered Jan 16 at 7:52
Robert ZRobert Z
101k1071144
$endgroup$
$begingroup$
Could you explain why It is from 0 to $pi$?
$endgroup$
– NPLS
Jan 16 at 7:56
$begingroup$
Draw your circle: it is centered in $(0,1)$ with radius $1$. It is completely contained in the upper half-plane and therefore the angle $theta$ varies in $[0,pi]$ (the ray starts in the origin in cylindrical coordinates).
$endgroup$
– Robert Z
Jan 16 at 8:00
$begingroup$
Thanks, you are magic!
$endgroup$
– NPLS
Jan 16 at 8:07
$begingroup$
@NPLS Fine! Let me know if you obtain the correct result.
$endgroup$
– Robert Z
Jan 16 at 8:12
$begingroup$
yup $10pi -32/3$
$endgroup$
– NPLS
Jan 16 at 8:55
|
show 7 more comments
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
You are almost correct. The angle $theta$ should be in the interval $[0,pi]$, so $0leq rleq 2sinthetageq 0$ (note that $r^2leq 2rsintheta$). Thus your triple integral should be
$$int_{0}^{pi}int_{0}^{2sin{theta}}int_{0}^{10-3r}rdzdrdtheta.$$
So what is the final result?
answered Jan 16 at 7:52
Robert ZRobert Z
101k1071144
$endgroup$
$begingroup$
Could you explain why It is from 0 to $pi$?
$endgroup$
– NPLS
Jan 16 at 7:56
$begingroup$
Draw your circle: it is centered in $(0,1)$ with radius $1$. It is completely contained in the upper half-plane and therefore the angle $theta$ varies in $[0,pi]$ (the ray starts in the origin in cylindrical coordinates).
$endgroup$
– Robert Z
Jan 16 at 8:00
$begingroup$
Thanks, you are magic!
$endgroup$
– NPLS
Jan 16 at 8:07
$begingroup$
@NPLS Fine! Let me know if you obtain the correct result.
$endgroup$
– Robert Z
Jan 16 at 8:12
$begingroup$
yup $10pi -32/3$
$endgroup$
– NPLS
Jan 16 at 8:55
|
show 7 more comments
$begingroup$
You are almost correct. The angle $theta$ should be in the interval $[0,pi]$, so $0leq rleq 2sinthetageq 0$ (note that $r^2leq 2rsintheta$). Thus your triple integral should be
$$int_{0}^{pi}int_{0}^{2sin{theta}}int_{0}^{10-3r}rdzdrdtheta.$$
So what is the final result?
answered Jan 16 at 7:52
Robert ZRobert Z
101k1071144
$endgroup$
$begingroup$
Could you explain why It is from 0 to $pi$?
$endgroup$
– NPLS
Jan 16 at 7:56
$begingroup$
Draw your circle: it is centered in $(0,1)$ with radius $1$. It is completely contained in the upper half-plane and therefore the angle $theta$ varies in $[0,pi]$ (the ray starts in the origin in cylindrical coordinates).
$endgroup$
– Robert Z
Jan 16 at 8:00
$begingroup$
Thanks, you are magic!
$endgroup$
– NPLS
Jan 16 at 8:07
$begingroup$
@NPLS Fine! Let me know if you obtain the correct result.
$endgroup$
– Robert Z
Jan 16 at 8:12
$begingroup$
yup $10pi -32/3$
$endgroup$
– NPLS
Jan 16 at 8:55
|
show 7 more comments
$begingroup$
You are almost correct. The angle $theta$ should be in the interval $[0,pi]$, so $0leq rleq 2sinthetageq 0$ (note that $r^2leq 2rsintheta$). Thus your triple integral should be
$$int_{0}^{pi}int_{0}^{2sin{theta}}int_{0}^{10-3r}rdzdrdtheta.$$
So what is the final result?
answered Jan 16 at 7:52
Robert ZRobert Z
101k1071144
$endgroup$
You are almost correct. The angle $theta$ should be in the interval $[0,pi]$, so $0leq rleq 2sinthetageq 0$ (note that $r^2leq 2rsintheta$). Thus your triple integral should be
$$int_{0}^{pi}int_{0}^{2sin{theta}}int_{0}^{10-3r}rdzdrdtheta.$$
So what is the final result?
answered Jan 16 at 7:52
Robert ZRobert Z
101k1071144
edited Jan 16 at 7:57
answered Jan 16 at 7:52
Robert ZRobert Z
101k1071144
answered Jan 16 at 7:52
Robert ZRobert Z
101k1071144
answered Jan 16 at 7:52
Robert ZRobert Z
101k1071144
101k1071144
$begingroup$
Could you explain why It is from 0 to $pi$?
$endgroup$
– NPLS
Jan 16 at 7:56
$begingroup$
Draw your circle: it is centered in $(0,1)$ with radius $1$. It is completely contained in the upper half-plane and therefore the angle $theta$ varies in $[0,pi]$ (the ray starts in the origin in cylindrical coordinates).
$endgroup$
– Robert Z
Jan 16 at 8:00
$begingroup$
Thanks, you are magic!
$endgroup$
– NPLS
Jan 16 at 8:07
$begingroup$
@NPLS Fine! Let me know if you obtain the correct result.
$endgroup$
– Robert Z
Jan 16 at 8:12
$begingroup$
yup $10pi -32/3$
$endgroup$
– NPLS
Jan 16 at 8:55
|
show 7 more comments
$begingroup$
Could you explain why It is from 0 to $pi$?
$endgroup$
– NPLS
Jan 16 at 7:56
$begingroup$
Draw your circle: it is centered in $(0,1)$ with radius $1$. It is completely contained in the upper half-plane and therefore the angle $theta$ varies in $[0,pi]$ (the ray starts in the origin in cylindrical coordinates).
$endgroup$
– Robert Z
Jan 16 at 8:00
$begingroup$
Thanks, you are magic!
$endgroup$
– NPLS
Jan 16 at 8:07
$begingroup$
@NPLS Fine! Let me know if you obtain the correct result.
$endgroup$
– Robert Z
Jan 16 at 8:12
$begingroup$
yup $10pi -32/3$
$endgroup$
– NPLS
Jan 16 at 8:55
$begingroup$
Could you explain why It is from 0 to $pi$?
$endgroup$
– NPLS
Jan 16 at 7:56
$begingroup$
Could you explain why It is from 0 to $pi$?
$endgroup$
– NPLS
Jan 16 at 7:56
$begingroup$
Draw your circle: it is centered in $(0,1)$ with radius $1$. It is completely contained in the upper half-plane and therefore the angle $theta$ varies in $[0,pi]$ (the ray starts in the origin in cylindrical coordinates).
$endgroup$
– Robert Z
Jan 16 at 8:00
$begingroup$
Draw your circle: it is centered in $(0,1)$ with radius $1$. It is completely contained in the upper half-plane and therefore the angle $theta$ varies in $[0,pi]$ (the ray starts in the origin in cylindrical coordinates).
$endgroup$
– Robert Z
Jan 16 at 8:00
$begingroup$
Thanks, you are magic!
$endgroup$
– NPLS
Jan 16 at 8:07
$begingroup$
Thanks, you are magic!
$endgroup$
– NPLS
Jan 16 at 8:07
$begingroup$
@NPLS Fine! Let me know if you obtain the correct result.
$endgroup$
– Robert Z
Jan 16 at 8:12
$begingroup$
@NPLS Fine! Let me know if you obtain the correct result.
$endgroup$
– Robert Z
Jan 16 at 8:12
$begingroup$
yup $10pi -32/3$
$endgroup$
– NPLS
Jan 16 at 8:55
$begingroup$
yup $10pi -32/3$
$endgroup$
– NPLS
Jan 16 at 8:55
|
show 7 more comments
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