Calculate this triple integral in cylindrical coordinates, the result is different with triple integral in...
$begingroup$
I want to calculate triple integral
begin{equation}intlimits_{-1}^{1}intlimits_{-sqrt{1-x^2}}^{sqrt{1-x^2}}intlimits_{x^2+y^2}^1 2z dzdydx.end{equation}
(the surface is $z=x^2+y^2$, $0leq zleq 1$.)
In MAPLE, I have to calculate it, and the result is
$$dfrac{2}{3}pi.$$
Now I want calculate the triple integral with cylindrical coordinates, become this
begin{equation}intlimits_{0}^{2pi}intlimits_{0}^{1}intlimits_{r}^1 2zr dzdrdtheta.end{equation}
begin{eqnarray}
intlimits_{0}^{2pi}intlimits_{0}^{1}intlimits_{r}^1 2zr dzdrdtheta
&=&
intlimits_{0}^{2pi}intlimits_{0}^{1}left[z^2rright]_r^1drdtheta\
&=&
intlimits_{0}^{2pi}intlimits_{0}^{1}left[r-r^3right]drdtheta\
&=&
intlimits_{0}^{2pi}left[dfrac{1}{2}r^2-dfrac{1}{4}r^4right]_0^1dtheta\
&=&
intlimits_{0}^{2pi}left[dfrac{1}{4}right]dtheta\
&=&left[dfrac{1}{4}thetaright]_0^{2pi}\
&=&dfrac{1}{2}pi.
end{eqnarray}
The results are different.
Am I wrong?
integration
asked Dec 11 '18 at 3:21
Ongky Denny WijayaOngky Denny Wijaya
3078
$endgroup$
add a comment |
$begingroup$
I want to calculate triple integral
begin{equation}intlimits_{-1}^{1}intlimits_{-sqrt{1-x^2}}^{sqrt{1-x^2}}intlimits_{x^2+y^2}^1 2z dzdydx.end{equation}
(the surface is $z=x^2+y^2$, $0leq zleq 1$.)
In MAPLE, I have to calculate it, and the result is
$$dfrac{2}{3}pi.$$
Now I want calculate the triple integral with cylindrical coordinates, become this
begin{equation}intlimits_{0}^{2pi}intlimits_{0}^{1}intlimits_{r}^1 2zr dzdrdtheta.end{equation}
begin{eqnarray}
intlimits_{0}^{2pi}intlimits_{0}^{1}intlimits_{r}^1 2zr dzdrdtheta
&=&
intlimits_{0}^{2pi}intlimits_{0}^{1}left[z^2rright]_r^1drdtheta\
&=&
intlimits_{0}^{2pi}intlimits_{0}^{1}left[r-r^3right]drdtheta\
&=&
intlimits_{0}^{2pi}left[dfrac{1}{2}r^2-dfrac{1}{4}r^4right]_0^1dtheta\
&=&
intlimits_{0}^{2pi}left[dfrac{1}{4}right]dtheta\
&=&left[dfrac{1}{4}thetaright]_0^{2pi}\
&=&dfrac{1}{2}pi.
end{eqnarray}
The results are different.
Am I wrong?
integration
asked Dec 11 '18 at 3:21
Ongky Denny WijayaOngky Denny Wijaya
3078
$endgroup$
add a comment |
$begingroup$
I want to calculate triple integral
begin{equation}intlimits_{-1}^{1}intlimits_{-sqrt{1-x^2}}^{sqrt{1-x^2}}intlimits_{x^2+y^2}^1 2z dzdydx.end{equation}
(the surface is $z=x^2+y^2$, $0leq zleq 1$.)
In MAPLE, I have to calculate it, and the result is
$$dfrac{2}{3}pi.$$
Now I want calculate the triple integral with cylindrical coordinates, become this
begin{equation}intlimits_{0}^{2pi}intlimits_{0}^{1}intlimits_{r}^1 2zr dzdrdtheta.end{equation}
begin{eqnarray}
intlimits_{0}^{2pi}intlimits_{0}^{1}intlimits_{r}^1 2zr dzdrdtheta
&=&
intlimits_{0}^{2pi}intlimits_{0}^{1}left[z^2rright]_r^1drdtheta\
&=&
intlimits_{0}^{2pi}intlimits_{0}^{1}left[r-r^3right]drdtheta\
&=&
intlimits_{0}^{2pi}left[dfrac{1}{2}r^2-dfrac{1}{4}r^4right]_0^1dtheta\
&=&
intlimits_{0}^{2pi}left[dfrac{1}{4}right]dtheta\
&=&left[dfrac{1}{4}thetaright]_0^{2pi}\
&=&dfrac{1}{2}pi.
end{eqnarray}
The results are different.
Am I wrong?
integration
asked Dec 11 '18 at 3:21
Ongky Denny WijayaOngky Denny Wijaya
3078
$endgroup$
I want to calculate triple integral
begin{equation}intlimits_{-1}^{1}intlimits_{-sqrt{1-x^2}}^{sqrt{1-x^2}}intlimits_{x^2+y^2}^1 2z dzdydx.end{equation}
(the surface is $z=x^2+y^2$, $0leq zleq 1$.)
In MAPLE, I have to calculate it, and the result is
$$dfrac{2}{3}pi.$$
Now I want calculate the triple integral with cylindrical coordinates, become this
begin{equation}intlimits_{0}^{2pi}intlimits_{0}^{1}intlimits_{r}^1 2zr dzdrdtheta.end{equation}
begin{eqnarray}
intlimits_{0}^{2pi}intlimits_{0}^{1}intlimits_{r}^1 2zr dzdrdtheta
&=&
intlimits_{0}^{2pi}intlimits_{0}^{1}left[z^2rright]_r^1drdtheta\
&=&
intlimits_{0}^{2pi}intlimits_{0}^{1}left[r-r^3right]drdtheta\
&=&
intlimits_{0}^{2pi}left[dfrac{1}{2}r^2-dfrac{1}{4}r^4right]_0^1dtheta\
&=&
intlimits_{0}^{2pi}left[dfrac{1}{4}right]dtheta\
&=&left[dfrac{1}{4}thetaright]_0^{2pi}\
&=&dfrac{1}{2}pi.
end{eqnarray}
The results are different.
Am I wrong?
integration
integration
asked Dec 11 '18 at 3:21
Ongky Denny WijayaOngky Denny Wijaya
3078
asked Dec 11 '18 at 3:21
Ongky Denny WijayaOngky Denny Wijaya
3078
asked Dec 11 '18 at 3:21
Ongky Denny WijayaOngky Denny Wijaya
3078
asked Dec 11 '18 at 3:21
Ongky Denny WijayaOngky Denny Wijaya
3078
asked Dec 11 '18 at 3:21
Ongky Denny WijayaOngky Denny Wijaya
3078
3078
add a comment |
add a comment |
1 Answer
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In cylindrical coordinate system, one has $$r^2 = x^2+y^2.$$
Hence, your integral should become:
$$int_0^{2pi}int_0^1int_{r^2}^12zrdzdrdtheta$$
answered Dec 11 '18 at 4:50
dezdichadodezdichado
6,3811929
$endgroup$
$begingroup$
why theta = 0 into theta =1?
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 4:56
$begingroup$
@ of course typo. edited
$endgroup$
– dezdichado
Dec 11 '18 at 5:03
1
$begingroup$
OK, I'm wrong in x^2+y^2, Now the answer is correct.
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 5:50
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
In cylindrical coordinate system, one has $$r^2 = x^2+y^2.$$
Hence, your integral should become:
$$int_0^{2pi}int_0^1int_{r^2}^12zrdzdrdtheta$$
answered Dec 11 '18 at 4:50
dezdichadodezdichado
6,3811929
$endgroup$
$begingroup$
why theta = 0 into theta =1?
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 4:56
$begingroup$
@ of course typo. edited
$endgroup$
– dezdichado
Dec 11 '18 at 5:03
1
$begingroup$
OK, I'm wrong in x^2+y^2, Now the answer is correct.
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 5:50
add a comment |
$begingroup$
In cylindrical coordinate system, one has $$r^2 = x^2+y^2.$$
Hence, your integral should become:
$$int_0^{2pi}int_0^1int_{r^2}^12zrdzdrdtheta$$
answered Dec 11 '18 at 4:50
dezdichadodezdichado
6,3811929
$endgroup$
$begingroup$
why theta = 0 into theta =1?
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 4:56
$begingroup$
@ of course typo. edited
$endgroup$
– dezdichado
Dec 11 '18 at 5:03
1
$begingroup$
OK, I'm wrong in x^2+y^2, Now the answer is correct.
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 5:50
add a comment |
$begingroup$
In cylindrical coordinate system, one has $$r^2 = x^2+y^2.$$
Hence, your integral should become:
$$int_0^{2pi}int_0^1int_{r^2}^12zrdzdrdtheta$$
answered Dec 11 '18 at 4:50
dezdichadodezdichado
6,3811929
$endgroup$
In cylindrical coordinate system, one has $$r^2 = x^2+y^2.$$
Hence, your integral should become:
$$int_0^{2pi}int_0^1int_{r^2}^12zrdzdrdtheta$$
answered Dec 11 '18 at 4:50
dezdichadodezdichado
6,3811929
edited Dec 11 '18 at 5:03
answered Dec 11 '18 at 4:50
dezdichadodezdichado
6,3811929
answered Dec 11 '18 at 4:50
dezdichadodezdichado
6,3811929
answered Dec 11 '18 at 4:50
dezdichadodezdichado
6,3811929
6,3811929
$begingroup$
why theta = 0 into theta =1?
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 4:56
$begingroup$
@ of course typo. edited
$endgroup$
– dezdichado
Dec 11 '18 at 5:03
1
$begingroup$
OK, I'm wrong in x^2+y^2, Now the answer is correct.
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 5:50
add a comment |
$begingroup$
why theta = 0 into theta =1?
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 4:56
$begingroup$
@ of course typo. edited
$endgroup$
– dezdichado
Dec 11 '18 at 5:03
1
$begingroup$
OK, I'm wrong in x^2+y^2, Now the answer is correct.
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 5:50
$begingroup$
why theta = 0 into theta =1?
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 4:56
$begingroup$
why theta = 0 into theta =1?
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 4:56
$begingroup$
@ of course typo. edited
$endgroup$
– dezdichado
Dec 11 '18 at 5:03
$begingroup$
@ of course typo. edited
$endgroup$
– dezdichado
Dec 11 '18 at 5:03
1
1
$begingroup$
OK, I'm wrong in x^2+y^2, Now the answer is correct.
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 5:50
$begingroup$
OK, I'm wrong in x^2+y^2, Now the answer is correct.
$endgroup$
– Ongky Denny Wijaya
Dec 11 '18 at 5:50
add a comment |
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