Prove that the volume of a cylinder is pi*a^2*h using triple integration and spherical coordinates
up vote
1
down vote
favorite
We have a cylinder of radius a and height h. We need to prove its volume to be equal to pi*a^2*h using triple integral and spherical coordinates. The best way to solve this problem is to divide the cylinder to two volume regions , the first region is the one defined by phi to range from 0 to arctan(a/h) and the second region is the one defined by phi to range from arctan(a/h) till pi/2. Surley , for both regions theta will vary from 0 till 2*pi. However , i tried many times to find the proper limits for the radius for each region but i failed. So how do i properly define the radius for each region ?
Edit : I know the limits of each region for the radius r= 0 till asecθ and r= 0 till bcscθ but my question exactly is how to drive them ?
integration spherical-coordinates
asked Nov 13 at 21:02
John adams
236
add a comment |
up vote
1
down vote
favorite
We have a cylinder of radius a and height h. We need to prove its volume to be equal to pi*a^2*h using triple integral and spherical coordinates. The best way to solve this problem is to divide the cylinder to two volume regions , the first region is the one defined by phi to range from 0 to arctan(a/h) and the second region is the one defined by phi to range from arctan(a/h) till pi/2. Surley , for both regions theta will vary from 0 till 2*pi. However , i tried many times to find the proper limits for the radius for each region but i failed. So how do i properly define the radius for each region ?
Edit : I know the limits of each region for the radius r= 0 till asecθ and r= 0 till bcscθ but my question exactly is how to drive them ?
integration spherical-coordinates
asked Nov 13 at 21:02
John adams
236
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 13 at 21:03
The equations for vertical and horizontal lines in polar coordinate are $r=a sec theta$ and $r=bcsc theta.$ You can adapt these readily.
– B. Goddard
Nov 13 at 21:07
@LordSharktheUnknown Ok
– John adams
Nov 13 at 21:08
@B.Goddard Please explain how do you drive them
– John adams
Nov 13 at 21:08
Start with $y=a$ is the same as $rsin theta = a.$
– B. Goddard
Nov 13 at 21:27
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We have a cylinder of radius a and height h. We need to prove its volume to be equal to pi*a^2*h using triple integral and spherical coordinates. The best way to solve this problem is to divide the cylinder to two volume regions , the first region is the one defined by phi to range from 0 to arctan(a/h) and the second region is the one defined by phi to range from arctan(a/h) till pi/2. Surley , for both regions theta will vary from 0 till 2*pi. However , i tried many times to find the proper limits for the radius for each region but i failed. So how do i properly define the radius for each region ?
Edit : I know the limits of each region for the radius r= 0 till asecθ and r= 0 till bcscθ but my question exactly is how to drive them ?
integration spherical-coordinates
asked Nov 13 at 21:02
John adams
236
We have a cylinder of radius a and height h. We need to prove its volume to be equal to pi*a^2*h using triple integral and spherical coordinates. The best way to solve this problem is to divide the cylinder to two volume regions , the first region is the one defined by phi to range from 0 to arctan(a/h) and the second region is the one defined by phi to range from arctan(a/h) till pi/2. Surley , for both regions theta will vary from 0 till 2*pi. However , i tried many times to find the proper limits for the radius for each region but i failed. So how do i properly define the radius for each region ?
Edit : I know the limits of each region for the radius r= 0 till asecθ and r= 0 till bcscθ but my question exactly is how to drive them ?
integration spherical-coordinates
integration spherical-coordinates
asked Nov 13 at 21:02
John adams
236
asked Nov 13 at 21:02
John adams
236
edited Nov 13 at 21:18
asked Nov 13 at 21:02
John adams
236
asked Nov 13 at 21:02
John adams
236
asked Nov 13 at 21:02
John adams
236
236
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 13 at 21:03
The equations for vertical and horizontal lines in polar coordinate are $r=a sec theta$ and $r=bcsc theta.$ You can adapt these readily.
– B. Goddard
Nov 13 at 21:07
@LordSharktheUnknown Ok
– John adams
Nov 13 at 21:08
@B.Goddard Please explain how do you drive them
– John adams
Nov 13 at 21:08
Start with $y=a$ is the same as $rsin theta = a.$
– B. Goddard
Nov 13 at 21:27
add a comment |
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 13 at 21:03
The equations for vertical and horizontal lines in polar coordinate are $r=a sec theta$ and $r=bcsc theta.$ You can adapt these readily.
– B. Goddard
Nov 13 at 21:07
@LordSharktheUnknown Ok
– John adams
Nov 13 at 21:08
@B.Goddard Please explain how do you drive them
– John adams
Nov 13 at 21:08
Start with $y=a$ is the same as $rsin theta = a.$
– B. Goddard
Nov 13 at 21:27
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 13 at 21:03
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 13 at 21:03
The equations for vertical and horizontal lines in polar coordinate are $r=a sec theta$ and $r=bcsc theta.$ You can adapt these readily.
– B. Goddard
Nov 13 at 21:07
The equations for vertical and horizontal lines in polar coordinate are $r=a sec theta$ and $r=bcsc theta.$ You can adapt these readily.
– B. Goddard
Nov 13 at 21:07
@LordSharktheUnknown Ok
– John adams
Nov 13 at 21:08
@LordSharktheUnknown Ok
– John adams
Nov 13 at 21:08
@B.Goddard Please explain how do you drive them
– John adams
Nov 13 at 21:08
@B.Goddard Please explain how do you drive them
– John adams
Nov 13 at 21:08
Start with $y=a$ is the same as $rsin theta = a.$
– B. Goddard
Nov 13 at 21:27
Start with $y=a$ is the same as $rsin theta = a.$
– B. Goddard
Nov 13 at 21:27
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
I would say that to find the volume of a cylinder use cylindrical coordinates.
But you want to use spherical.
Your boundaries
$x^2 + y^2 = a^2\
z = 0\
z = h$
$x = rhocosthetasinphi\
y = rhosinthetasinphi\
z = rhocos phi$
$rho^2cos^2 thetasin^2phi + rho^2sin^2 thetasin^2phi = a^2\
rho^2sin^2phi = a^2\
rho = acscphi$
$rhocos phi = h\
rho = hsec phi$
$acscphi = hsec phi\
tanphi = frac ah$
$int_limits0^{2pi}int_limits0^{arctan frac {a}{h}}int_limits0^{hsec phi} rho^2sin phi drho dphi dtheta\ + int_limits0^{2pi}int_limits{arctan frac {a}{h}}^frac {pi}{2}int_limits0^{acsc phi} rho^2sin phi drho dphi dtheta$
answered Nov 13 at 21:14
Doug M
42.6k31752
How did you drive the limits of upper limits of the radius for each region ?
– John adams
Nov 13 at 21:17
@Johnadams Did you ask this question before I completed my last edit? Is it all clear now?
– Doug M
Nov 13 at 21:29
Yes , Thank you very much.
– John adams
Nov 13 at 21:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I would say that to find the volume of a cylinder use cylindrical coordinates.
But you want to use spherical.
Your boundaries
$x^2 + y^2 = a^2\
z = 0\
z = h$
$x = rhocosthetasinphi\
y = rhosinthetasinphi\
z = rhocos phi$
$rho^2cos^2 thetasin^2phi + rho^2sin^2 thetasin^2phi = a^2\
rho^2sin^2phi = a^2\
rho = acscphi$
$rhocos phi = h\
rho = hsec phi$
$acscphi = hsec phi\
tanphi = frac ah$
$int_limits0^{2pi}int_limits0^{arctan frac {a}{h}}int_limits0^{hsec phi} rho^2sin phi drho dphi dtheta\ + int_limits0^{2pi}int_limits{arctan frac {a}{h}}^frac {pi}{2}int_limits0^{acsc phi} rho^2sin phi drho dphi dtheta$
answered Nov 13 at 21:14
Doug M
42.6k31752
How did you drive the limits of upper limits of the radius for each region ?
– John adams
Nov 13 at 21:17
@Johnadams Did you ask this question before I completed my last edit? Is it all clear now?
– Doug M
Nov 13 at 21:29
Yes , Thank you very much.
– John adams
Nov 13 at 21:30
add a comment |
up vote
1
down vote
accepted
I would say that to find the volume of a cylinder use cylindrical coordinates.
But you want to use spherical.
Your boundaries
$x^2 + y^2 = a^2\
z = 0\
z = h$
$x = rhocosthetasinphi\
y = rhosinthetasinphi\
z = rhocos phi$
$rho^2cos^2 thetasin^2phi + rho^2sin^2 thetasin^2phi = a^2\
rho^2sin^2phi = a^2\
rho = acscphi$
$rhocos phi = h\
rho = hsec phi$
$acscphi = hsec phi\
tanphi = frac ah$
$int_limits0^{2pi}int_limits0^{arctan frac {a}{h}}int_limits0^{hsec phi} rho^2sin phi drho dphi dtheta\ + int_limits0^{2pi}int_limits{arctan frac {a}{h}}^frac {pi}{2}int_limits0^{acsc phi} rho^2sin phi drho dphi dtheta$
answered Nov 13 at 21:14
Doug M
42.6k31752
How did you drive the limits of upper limits of the radius for each region ?
– John adams
Nov 13 at 21:17
@Johnadams Did you ask this question before I completed my last edit? Is it all clear now?
– Doug M
Nov 13 at 21:29
Yes , Thank you very much.
– John adams
Nov 13 at 21:30
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I would say that to find the volume of a cylinder use cylindrical coordinates.
But you want to use spherical.
Your boundaries
$x^2 + y^2 = a^2\
z = 0\
z = h$
$x = rhocosthetasinphi\
y = rhosinthetasinphi\
z = rhocos phi$
$rho^2cos^2 thetasin^2phi + rho^2sin^2 thetasin^2phi = a^2\
rho^2sin^2phi = a^2\
rho = acscphi$
$rhocos phi = h\
rho = hsec phi$
$acscphi = hsec phi\
tanphi = frac ah$
$int_limits0^{2pi}int_limits0^{arctan frac {a}{h}}int_limits0^{hsec phi} rho^2sin phi drho dphi dtheta\ + int_limits0^{2pi}int_limits{arctan frac {a}{h}}^frac {pi}{2}int_limits0^{acsc phi} rho^2sin phi drho dphi dtheta$
answered Nov 13 at 21:14
Doug M
42.6k31752
I would say that to find the volume of a cylinder use cylindrical coordinates.
But you want to use spherical.
Your boundaries
$x^2 + y^2 = a^2\
z = 0\
z = h$
$x = rhocosthetasinphi\
y = rhosinthetasinphi\
z = rhocos phi$
$rho^2cos^2 thetasin^2phi + rho^2sin^2 thetasin^2phi = a^2\
rho^2sin^2phi = a^2\
rho = acscphi$
$rhocos phi = h\
rho = hsec phi$
$acscphi = hsec phi\
tanphi = frac ah$
$int_limits0^{2pi}int_limits0^{arctan frac {a}{h}}int_limits0^{hsec phi} rho^2sin phi drho dphi dtheta\ + int_limits0^{2pi}int_limits{arctan frac {a}{h}}^frac {pi}{2}int_limits0^{acsc phi} rho^2sin phi drho dphi dtheta$
answered Nov 13 at 21:14
Doug M
42.6k31752
edited Nov 13 at 21:20
answered Nov 13 at 21:14
Doug M
42.6k31752
answered Nov 13 at 21:14
Doug M
42.6k31752
answered Nov 13 at 21:14
Doug M
42.6k31752
42.6k31752
How did you drive the limits of upper limits of the radius for each region ?
– John adams
Nov 13 at 21:17
@Johnadams Did you ask this question before I completed my last edit? Is it all clear now?
– Doug M
Nov 13 at 21:29
Yes , Thank you very much.
– John adams
Nov 13 at 21:30
add a comment |
How did you drive the limits of upper limits of the radius for each region ?
– John adams
Nov 13 at 21:17
@Johnadams Did you ask this question before I completed my last edit? Is it all clear now?
– Doug M
Nov 13 at 21:29
Yes , Thank you very much.
– John adams
Nov 13 at 21:30
How did you drive the limits of upper limits of the radius for each region ?
– John adams
Nov 13 at 21:17
How did you drive the limits of upper limits of the radius for each region ?
– John adams
Nov 13 at 21:17
@Johnadams Did you ask this question before I completed my last edit? Is it all clear now?
– Doug M
Nov 13 at 21:29
@Johnadams Did you ask this question before I completed my last edit? Is it all clear now?
– Doug M
Nov 13 at 21:29
Yes , Thank you very much.
– John adams
Nov 13 at 21:30
Yes , Thank you very much.
– John adams
Nov 13 at 21:30
add a comment |
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%2f2997332%2fprove-that-the-volume-of-a-cylinder-is-pia2h-using-triple-integration-and-sph%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
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 13 at 21:03
The equations for vertical and horizontal lines in polar coordinate are $r=a sec theta$ and $r=bcsc theta.$ You can adapt these readily.
– B. Goddard
Nov 13 at 21:07
@LordSharktheUnknown Ok
– John adams
Nov 13 at 21:08
@B.Goddard Please explain how do you drive them
– John adams
Nov 13 at 21:08
Start with $y=a$ is the same as $rsin theta = a.$
– B. Goddard
Nov 13 at 21:27