Triple integrals (Find volume): The solid bounded by the sphere $r = 2 cos$ $ phi $ and the hemisphere $r =...
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Here is the exact question:
https://imgur.com/a/cBQC8su!
My particular question regards the range of $phi$; $phi$ certainly lives between $0$ $le$ $phi$ $le$ $frac {pi}{2}$.
$rho = 1$ intersects with $ rho = 2cosphi$ at $frac {pi}{3}$.
I thought that the range of $phi$ would be $0$ $le$ $phi$ $le$ $frac {pi}{2}$, where $0$ $le$ $phi$ $le$ $frac {pi}{3}$ to $frac {pi}{3}$ $le$ $phi$ $le$ $frac {pi}{2}$. Any helpful tips?
The book lists the answer as $frac {5pi}{12}$.
definite-integrals spherical-coordinates
asked Nov 18 at 9:11
Michael Ramage
61
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up vote
1
down vote
favorite
Here is the exact question:
https://imgur.com/a/cBQC8su!
My particular question regards the range of $phi$; $phi$ certainly lives between $0$ $le$ $phi$ $le$ $frac {pi}{2}$.
$rho = 1$ intersects with $ rho = 2cosphi$ at $frac {pi}{3}$.
I thought that the range of $phi$ would be $0$ $le$ $phi$ $le$ $frac {pi}{2}$, where $0$ $le$ $phi$ $le$ $frac {pi}{3}$ to $frac {pi}{3}$ $le$ $phi$ $le$ $frac {pi}{2}$. Any helpful tips?
The book lists the answer as $frac {5pi}{12}$.
definite-integrals spherical-coordinates
asked Nov 18 at 9:11
Michael Ramage
61
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is the exact question:
https://imgur.com/a/cBQC8su!
My particular question regards the range of $phi$; $phi$ certainly lives between $0$ $le$ $phi$ $le$ $frac {pi}{2}$.
$rho = 1$ intersects with $ rho = 2cosphi$ at $frac {pi}{3}$.
I thought that the range of $phi$ would be $0$ $le$ $phi$ $le$ $frac {pi}{2}$, where $0$ $le$ $phi$ $le$ $frac {pi}{3}$ to $frac {pi}{3}$ $le$ $phi$ $le$ $frac {pi}{2}$. Any helpful tips?
The book lists the answer as $frac {5pi}{12}$.
definite-integrals spherical-coordinates
asked Nov 18 at 9:11
Michael Ramage
61
Here is the exact question:
https://imgur.com/a/cBQC8su!
My particular question regards the range of $phi$; $phi$ certainly lives between $0$ $le$ $phi$ $le$ $frac {pi}{2}$.
$rho = 1$ intersects with $ rho = 2cosphi$ at $frac {pi}{3}$.
I thought that the range of $phi$ would be $0$ $le$ $phi$ $le$ $frac {pi}{2}$, where $0$ $le$ $phi$ $le$ $frac {pi}{3}$ to $frac {pi}{3}$ $le$ $phi$ $le$ $frac {pi}{2}$. Any helpful tips?
The book lists the answer as $frac {5pi}{12}$.
definite-integrals spherical-coordinates
definite-integrals spherical-coordinates
asked Nov 18 at 9:11
Michael Ramage
61
asked Nov 18 at 9:11
Michael Ramage
61
asked Nov 18 at 9:11
Michael Ramage
61
asked Nov 18 at 9:11
Michael Ramage
61
asked Nov 18 at 9:11
Michael Ramage
61
61
add a comment |
add a comment |
1 Answer
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The intersection $S$ of the two unit balls is a lens shaped object whose volume can easily be computed by the "washer method". One obtains
$${rm vol}(S)=2cdotint_{1/2}^1pi(1-r^2)>dr={5piover12} .$$
answered Nov 18 at 13:24
Christian Blatter
171k7111325
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The intersection $S$ of the two unit balls is a lens shaped object whose volume can easily be computed by the "washer method". One obtains
$${rm vol}(S)=2cdotint_{1/2}^1pi(1-r^2)>dr={5piover12} .$$
answered Nov 18 at 13:24
Christian Blatter
171k7111325
add a comment |
up vote
0
down vote
The intersection $S$ of the two unit balls is a lens shaped object whose volume can easily be computed by the "washer method". One obtains
$${rm vol}(S)=2cdotint_{1/2}^1pi(1-r^2)>dr={5piover12} .$$
answered Nov 18 at 13:24
Christian Blatter
171k7111325
add a comment |
up vote
0
down vote
up vote
0
down vote
The intersection $S$ of the two unit balls is a lens shaped object whose volume can easily be computed by the "washer method". One obtains
$${rm vol}(S)=2cdotint_{1/2}^1pi(1-r^2)>dr={5piover12} .$$
answered Nov 18 at 13:24
Christian Blatter
171k7111325
The intersection $S$ of the two unit balls is a lens shaped object whose volume can easily be computed by the "washer method". One obtains
$${rm vol}(S)=2cdotint_{1/2}^1pi(1-r^2)>dr={5piover12} .$$
answered Nov 18 at 13:24
Christian Blatter
171k7111325
answered Nov 18 at 13:24
Christian Blatter
171k7111325
answered Nov 18 at 13:24
Christian Blatter
171k7111325
answered Nov 18 at 13:24
Christian Blatter
171k7111325
171k7111325
add a comment |
add a comment |
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