Volume inside sphere bounded by plane in double integrals
up vote
0
down vote
favorite
I am trying to solve for volume below a plane bounded by a sphere given by
$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be
$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!
EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.
integration multivariable-calculus polar-coordinates
asked Nov 20 at 15:19
mstro
11
add a comment |
up vote
0
down vote
favorite
I am trying to solve for volume below a plane bounded by a sphere given by
$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be
$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!
EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.
integration multivariable-calculus polar-coordinates
asked Nov 20 at 15:19
mstro
11
Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 at 16:27
I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 at 8:36
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve for volume below a plane bounded by a sphere given by
$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be
$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!
EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.
integration multivariable-calculus polar-coordinates
asked Nov 20 at 15:19
mstro
11
I am trying to solve for volume below a plane bounded by a sphere given by
$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be
$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!
EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.
integration multivariable-calculus polar-coordinates
integration multivariable-calculus polar-coordinates
asked Nov 20 at 15:19
mstro
11
asked Nov 20 at 15:19
mstro
11
edited Nov 22 at 8:41
asked Nov 20 at 15:19
mstro
11
asked Nov 20 at 15:19
mstro
11
asked Nov 20 at 15:19
mstro
11
11
Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 at 16:27
I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 at 8:36
add a comment |
Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 at 16:27
I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 at 8:36
Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 at 16:27
Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 at 16:27
I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 at 8:36
I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 at 8:36
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
$$frac{1}{3}pi(9-c^2)c$$
The second is the volume between the cone and z = 0 which can be obtained by
$$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
Then the total volume is
$$pi(9c - frac{c^3}{3})$$
In fact it is easier to use xyz co-ordinates.
$$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$
answered Nov 21 at 22:47
KY Tang
12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
$$frac{1}{3}pi(9-c^2)c$$
The second is the volume between the cone and z = 0 which can be obtained by
$$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
Then the total volume is
$$pi(9c - frac{c^3}{3})$$
In fact it is easier to use xyz co-ordinates.
$$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$
answered Nov 21 at 22:47
KY Tang
12
add a comment |
up vote
0
down vote
The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
$$frac{1}{3}pi(9-c^2)c$$
The second is the volume between the cone and z = 0 which can be obtained by
$$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
Then the total volume is
$$pi(9c - frac{c^3}{3})$$
In fact it is easier to use xyz co-ordinates.
$$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$
answered Nov 21 at 22:47
KY Tang
12
add a comment |
up vote
0
down vote
up vote
0
down vote
The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
$$frac{1}{3}pi(9-c^2)c$$
The second is the volume between the cone and z = 0 which can be obtained by
$$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
Then the total volume is
$$pi(9c - frac{c^3}{3})$$
In fact it is easier to use xyz co-ordinates.
$$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$
answered Nov 21 at 22:47
KY Tang
12
The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
$$frac{1}{3}pi(9-c^2)c$$
The second is the volume between the cone and z = 0 which can be obtained by
$$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
Then the total volume is
$$pi(9c - frac{c^3}{3})$$
In fact it is easier to use xyz co-ordinates.
$$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$
answered Nov 21 at 22:47
KY Tang
12
answered Nov 21 at 22:47
KY Tang
12
answered Nov 21 at 22:47
KY Tang
12
answered Nov 21 at 22:47
KY Tang
12
12
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006445%2fvolume-inside-sphere-bounded-by-plane-in-double-integrals%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 at 16:27
I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 at 8:36