How to find the spherical coordinates limits calculate the volume of the solid region?
$begingroup$
The solid bounded below by the hemisphere $rho=1,$ $ 0leq z$ and above by the cardoid of revolution $rho=1+cosphi$
I am new to the triple integral in spherical coordinates.
I know that limits of $rho$ will be $1leqrholeq 1+cosphi$, the limits on $theta$ will be $0leqthetaleq2pi$. So the integral will be
$$int_{0}^{2pi}int_{?}^{?}int_{1}^{1+cosphi}(rho)^2sinphi drho dphi dtheta$$
But how to find the limits of $phi$ in this case, I don't know.
Can anyone help!
Thanks in advance!
multivariable-calculus multiple-integral
edited Jan 7 at 17:29
Shubham Johri
4,885717
asked Jan 7 at 17:05
Noor AslamNoor Aslam
15112
$endgroup$
add a comment |
$begingroup$
The solid bounded below by the hemisphere $rho=1,$ $ 0leq z$ and above by the cardoid of revolution $rho=1+cosphi$
I am new to the triple integral in spherical coordinates.
I know that limits of $rho$ will be $1leqrholeq 1+cosphi$, the limits on $theta$ will be $0leqthetaleq2pi$. So the integral will be
$$int_{0}^{2pi}int_{?}^{?}int_{1}^{1+cosphi}(rho)^2sinphi drho dphi dtheta$$
But how to find the limits of $phi$ in this case, I don't know.
Can anyone help!
Thanks in advance!
multivariable-calculus multiple-integral
edited Jan 7 at 17:29
Shubham Johri
4,885717
asked Jan 7 at 17:05
Noor AslamNoor Aslam
15112
$endgroup$
add a comment |
$begingroup$
The solid bounded below by the hemisphere $rho=1,$ $ 0leq z$ and above by the cardoid of revolution $rho=1+cosphi$
I am new to the triple integral in spherical coordinates.
I know that limits of $rho$ will be $1leqrholeq 1+cosphi$, the limits on $theta$ will be $0leqthetaleq2pi$. So the integral will be
$$int_{0}^{2pi}int_{?}^{?}int_{1}^{1+cosphi}(rho)^2sinphi drho dphi dtheta$$
But how to find the limits of $phi$ in this case, I don't know.
Can anyone help!
Thanks in advance!
multivariable-calculus multiple-integral
edited Jan 7 at 17:29
Shubham Johri
4,885717
asked Jan 7 at 17:05
Noor AslamNoor Aslam
15112
$endgroup$
The solid bounded below by the hemisphere $rho=1,$ $ 0leq z$ and above by the cardoid of revolution $rho=1+cosphi$
I am new to the triple integral in spherical coordinates.
I know that limits of $rho$ will be $1leqrholeq 1+cosphi$, the limits on $theta$ will be $0leqthetaleq2pi$. So the integral will be
$$int_{0}^{2pi}int_{?}^{?}int_{1}^{1+cosphi}(rho)^2sinphi drho dphi dtheta$$
But how to find the limits of $phi$ in this case, I don't know.
Can anyone help!
Thanks in advance!
multivariable-calculus multiple-integral
multivariable-calculus multiple-integral
edited Jan 7 at 17:29
Shubham Johri
4,885717
asked Jan 7 at 17:05
Noor AslamNoor Aslam
15112
edited Jan 7 at 17:29
Shubham Johri
4,885717
asked Jan 7 at 17:05
Noor AslamNoor Aslam
15112
edited Jan 7 at 17:29
Shubham Johri
4,885717
edited Jan 7 at 17:29
Shubham Johri
4,885717
edited Jan 7 at 17:29
Shubham Johri
4,885717
4,885717
asked Jan 7 at 17:05
Noor AslamNoor Aslam
15112
asked Jan 7 at 17:05
Noor AslamNoor Aslam
15112
asked Jan 7 at 17:05
Noor AslamNoor Aslam
15112
15112
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Notice that for the cardioid of revolution, $rhoge1$ for $0lephilepi/2$ and $rho<1$ for $pi/2<philepi$. It intersects the hemisphere only in the $xy$ plane. Therefore, $0lephilepi/2$.
answered Jan 7 at 18:16
Shubham JohriShubham Johri
4,885717
$endgroup$
$begingroup$
Sir the limits on the $rho$ and $theta$ I have written are correct?
$endgroup$
– Noor Aslam
Jan 8 at 12:47
$begingroup$
@NoorAslam They are correct
$endgroup$
– Shubham Johri
Jan 8 at 18:01
add a comment |
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1 Answer
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active
oldest
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1 Answer
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oldest
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votes
$begingroup$
Notice that for the cardioid of revolution, $rhoge1$ for $0lephilepi/2$ and $rho<1$ for $pi/2<philepi$. It intersects the hemisphere only in the $xy$ plane. Therefore, $0lephilepi/2$.
answered Jan 7 at 18:16
Shubham JohriShubham Johri
4,885717
$endgroup$
$begingroup$
Sir the limits on the $rho$ and $theta$ I have written are correct?
$endgroup$
– Noor Aslam
Jan 8 at 12:47
$begingroup$
@NoorAslam They are correct
$endgroup$
– Shubham Johri
Jan 8 at 18:01
add a comment |
$begingroup$
Notice that for the cardioid of revolution, $rhoge1$ for $0lephilepi/2$ and $rho<1$ for $pi/2<philepi$. It intersects the hemisphere only in the $xy$ plane. Therefore, $0lephilepi/2$.
answered Jan 7 at 18:16
Shubham JohriShubham Johri
4,885717
$endgroup$
$begingroup$
Sir the limits on the $rho$ and $theta$ I have written are correct?
$endgroup$
– Noor Aslam
Jan 8 at 12:47
$begingroup$
@NoorAslam They are correct
$endgroup$
– Shubham Johri
Jan 8 at 18:01
add a comment |
$begingroup$
Notice that for the cardioid of revolution, $rhoge1$ for $0lephilepi/2$ and $rho<1$ for $pi/2<philepi$. It intersects the hemisphere only in the $xy$ plane. Therefore, $0lephilepi/2$.
answered Jan 7 at 18:16
Shubham JohriShubham Johri
4,885717
$endgroup$
Notice that for the cardioid of revolution, $rhoge1$ for $0lephilepi/2$ and $rho<1$ for $pi/2<philepi$. It intersects the hemisphere only in the $xy$ plane. Therefore, $0lephilepi/2$.
answered Jan 7 at 18:16
Shubham JohriShubham Johri
4,885717
answered Jan 7 at 18:16
Shubham JohriShubham Johri
4,885717
answered Jan 7 at 18:16
Shubham JohriShubham Johri
4,885717
answered Jan 7 at 18:16
Shubham JohriShubham Johri
4,885717
4,885717
$begingroup$
Sir the limits on the $rho$ and $theta$ I have written are correct?
$endgroup$
– Noor Aslam
Jan 8 at 12:47
$begingroup$
@NoorAslam They are correct
$endgroup$
– Shubham Johri
Jan 8 at 18:01
add a comment |
$begingroup$
Sir the limits on the $rho$ and $theta$ I have written are correct?
$endgroup$
– Noor Aslam
Jan 8 at 12:47
$begingroup$
@NoorAslam They are correct
$endgroup$
– Shubham Johri
Jan 8 at 18:01
$begingroup$
Sir the limits on the $rho$ and $theta$ I have written are correct?
$endgroup$
– Noor Aslam
Jan 8 at 12:47
$begingroup$
Sir the limits on the $rho$ and $theta$ I have written are correct?
$endgroup$
– Noor Aslam
Jan 8 at 12:47
$begingroup$
@NoorAslam They are correct
$endgroup$
– Shubham Johri
Jan 8 at 18:01
$begingroup$
@NoorAslam They are correct
$endgroup$
– Shubham Johri
Jan 8 at 18:01
add a comment |
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