How to calculate this surface area? (portion of a cylinder inside a sphere )
up vote
4
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The surface area of the portion of the cylinder $x^2+y^2=8y$ located inside of the sphere $x^2+y^2+z^2=64$
I'm stuck, so any tip will be helpful
Thanks in advance!
multivariable-calculus
asked May 22 '17 at 19:17
user418360
365
add a comment |
up vote
4
down vote
favorite
The surface area of the portion of the cylinder $x^2+y^2=8y$ located inside of the sphere $x^2+y^2+z^2=64$
I'm stuck, so any tip will be helpful
Thanks in advance!
multivariable-calculus
asked May 22 '17 at 19:17
user418360
365
By area do you mean surface area?
– theyaoster
May 22 '17 at 19:30
yes, I do! surface area
– user418360
May 22 '17 at 19:31
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The surface area of the portion of the cylinder $x^2+y^2=8y$ located inside of the sphere $x^2+y^2+z^2=64$
I'm stuck, so any tip will be helpful
Thanks in advance!
multivariable-calculus
asked May 22 '17 at 19:17
user418360
365
The surface area of the portion of the cylinder $x^2+y^2=8y$ located inside of the sphere $x^2+y^2+z^2=64$
I'm stuck, so any tip will be helpful
Thanks in advance!
multivariable-calculus
multivariable-calculus
asked May 22 '17 at 19:17
user418360
365
asked May 22 '17 at 19:17
user418360
365
edited May 22 '17 at 19:30
asked May 22 '17 at 19:17
user418360
365
asked May 22 '17 at 19:17
user418360
365
asked May 22 '17 at 19:17
user418360
365
365
By area do you mean surface area?
– theyaoster
May 22 '17 at 19:30
yes, I do! surface area
– user418360
May 22 '17 at 19:31
add a comment |
By area do you mean surface area?
– theyaoster
May 22 '17 at 19:30
yes, I do! surface area
– user418360
May 22 '17 at 19:31
By area do you mean surface area?
– theyaoster
May 22 '17 at 19:30
By area do you mean surface area?
– theyaoster
May 22 '17 at 19:30
yes, I do! surface area
– user418360
May 22 '17 at 19:31
yes, I do! surface area
– user418360
May 22 '17 at 19:31
add a comment |
1 Answer
1
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oldest
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0
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accepted
$A = iint dS$
$S: x^2 + y^2 = 8y$
Convert to cylindrical.
$x = rcostheta\
y = rsintheta\
z = z$
Plug these into the equation of the cylinder.
$r = 8sintheta$
And substitute back for parameterization of the surface
$x = 8sinthetacostheta = 4sin 2theta\
y = 8sin^2theta = 4 - 4cos 2theta\
z = z$
$dS = $$|(frac {partial x}{partial theta}, frac {partial y}{partial theta},frac {partial z}{partial theta})times (frac {partial x}{partial z}, frac {partial y}{partial z},frac {partial z}{partial z})|\
|(8cos2theta, 8sin 2theta, 0) times (0,0,1)| = |(8sin2theta, -8cos 2theta, 0)| = 8 dz dtheta$
$iint 8 dz dtheta$
The sphere will establish the limits for z.
$16sin^2 2theta + 16 -32cos 2theta + 16cos^2 2theta + z^2 = 64\
z^2 = 32 + 32cos 2theta = 64cos^2theta$
$2int_0^{pi}int_0^{8costheta} 8 dz dtheta$
edited Nov 19 at 21:42
Matheus Ramos
32
answered May 22 '17 at 19:36
Doug M
43.5k31854
But I need the cylinder surface area, this would be the surface sphere area
– user418360
May 22 '17 at 19:41
That is better...
– Doug M
May 22 '17 at 20:00
Thank you so much! it's perfect.
– user418360
May 22 '17 at 20:06
Can you explain the bounds?
– ifly6
Dec 5 at 16:11
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$A = iint dS$
$S: x^2 + y^2 = 8y$
Convert to cylindrical.
$x = rcostheta\
y = rsintheta\
z = z$
Plug these into the equation of the cylinder.
$r = 8sintheta$
And substitute back for parameterization of the surface
$x = 8sinthetacostheta = 4sin 2theta\
y = 8sin^2theta = 4 - 4cos 2theta\
z = z$
$dS = $$|(frac {partial x}{partial theta}, frac {partial y}{partial theta},frac {partial z}{partial theta})times (frac {partial x}{partial z}, frac {partial y}{partial z},frac {partial z}{partial z})|\
|(8cos2theta, 8sin 2theta, 0) times (0,0,1)| = |(8sin2theta, -8cos 2theta, 0)| = 8 dz dtheta$
$iint 8 dz dtheta$
The sphere will establish the limits for z.
$16sin^2 2theta + 16 -32cos 2theta + 16cos^2 2theta + z^2 = 64\
z^2 = 32 + 32cos 2theta = 64cos^2theta$
$2int_0^{pi}int_0^{8costheta} 8 dz dtheta$
edited Nov 19 at 21:42
Matheus Ramos
32
answered May 22 '17 at 19:36
Doug M
43.5k31854
But I need the cylinder surface area, this would be the surface sphere area
– user418360
May 22 '17 at 19:41
That is better...
– Doug M
May 22 '17 at 20:00
Thank you so much! it's perfect.
– user418360
May 22 '17 at 20:06
Can you explain the bounds?
– ifly6
Dec 5 at 16:11
add a comment |
up vote
0
down vote
accepted
$A = iint dS$
$S: x^2 + y^2 = 8y$
Convert to cylindrical.
$x = rcostheta\
y = rsintheta\
z = z$
Plug these into the equation of the cylinder.
$r = 8sintheta$
And substitute back for parameterization of the surface
$x = 8sinthetacostheta = 4sin 2theta\
y = 8sin^2theta = 4 - 4cos 2theta\
z = z$
$dS = $$|(frac {partial x}{partial theta}, frac {partial y}{partial theta},frac {partial z}{partial theta})times (frac {partial x}{partial z}, frac {partial y}{partial z},frac {partial z}{partial z})|\
|(8cos2theta, 8sin 2theta, 0) times (0,0,1)| = |(8sin2theta, -8cos 2theta, 0)| = 8 dz dtheta$
$iint 8 dz dtheta$
The sphere will establish the limits for z.
$16sin^2 2theta + 16 -32cos 2theta + 16cos^2 2theta + z^2 = 64\
z^2 = 32 + 32cos 2theta = 64cos^2theta$
$2int_0^{pi}int_0^{8costheta} 8 dz dtheta$
edited Nov 19 at 21:42
Matheus Ramos
32
answered May 22 '17 at 19:36
Doug M
43.5k31854
But I need the cylinder surface area, this would be the surface sphere area
– user418360
May 22 '17 at 19:41
That is better...
– Doug M
May 22 '17 at 20:00
Thank you so much! it's perfect.
– user418360
May 22 '17 at 20:06
Can you explain the bounds?
– ifly6
Dec 5 at 16:11
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$A = iint dS$
$S: x^2 + y^2 = 8y$
Convert to cylindrical.
$x = rcostheta\
y = rsintheta\
z = z$
Plug these into the equation of the cylinder.
$r = 8sintheta$
And substitute back for parameterization of the surface
$x = 8sinthetacostheta = 4sin 2theta\
y = 8sin^2theta = 4 - 4cos 2theta\
z = z$
$dS = $$|(frac {partial x}{partial theta}, frac {partial y}{partial theta},frac {partial z}{partial theta})times (frac {partial x}{partial z}, frac {partial y}{partial z},frac {partial z}{partial z})|\
|(8cos2theta, 8sin 2theta, 0) times (0,0,1)| = |(8sin2theta, -8cos 2theta, 0)| = 8 dz dtheta$
$iint 8 dz dtheta$
The sphere will establish the limits for z.
$16sin^2 2theta + 16 -32cos 2theta + 16cos^2 2theta + z^2 = 64\
z^2 = 32 + 32cos 2theta = 64cos^2theta$
$2int_0^{pi}int_0^{8costheta} 8 dz dtheta$
edited Nov 19 at 21:42
Matheus Ramos
32
answered May 22 '17 at 19:36
Doug M
43.5k31854
$A = iint dS$
$S: x^2 + y^2 = 8y$
Convert to cylindrical.
$x = rcostheta\
y = rsintheta\
z = z$
Plug these into the equation of the cylinder.
$r = 8sintheta$
And substitute back for parameterization of the surface
$x = 8sinthetacostheta = 4sin 2theta\
y = 8sin^2theta = 4 - 4cos 2theta\
z = z$
$dS = $$|(frac {partial x}{partial theta}, frac {partial y}{partial theta},frac {partial z}{partial theta})times (frac {partial x}{partial z}, frac {partial y}{partial z},frac {partial z}{partial z})|\
|(8cos2theta, 8sin 2theta, 0) times (0,0,1)| = |(8sin2theta, -8cos 2theta, 0)| = 8 dz dtheta$
$iint 8 dz dtheta$
The sphere will establish the limits for z.
$16sin^2 2theta + 16 -32cos 2theta + 16cos^2 2theta + z^2 = 64\
z^2 = 32 + 32cos 2theta = 64cos^2theta$
$2int_0^{pi}int_0^{8costheta} 8 dz dtheta$
edited Nov 19 at 21:42
Matheus Ramos
32
answered May 22 '17 at 19:36
Doug M
43.5k31854
edited Nov 19 at 21:42
Matheus Ramos
32
edited Nov 19 at 21:42
Matheus Ramos
32
edited Nov 19 at 21:42
Matheus Ramos
32
32
answered May 22 '17 at 19:36
Doug M
43.5k31854
answered May 22 '17 at 19:36
Doug M
43.5k31854
answered May 22 '17 at 19:36
Doug M
43.5k31854
43.5k31854
But I need the cylinder surface area, this would be the surface sphere area
– user418360
May 22 '17 at 19:41
That is better...
– Doug M
May 22 '17 at 20:00
Thank you so much! it's perfect.
– user418360
May 22 '17 at 20:06
Can you explain the bounds?
– ifly6
Dec 5 at 16:11
add a comment |
But I need the cylinder surface area, this would be the surface sphere area
– user418360
May 22 '17 at 19:41
That is better...
– Doug M
May 22 '17 at 20:00
Thank you so much! it's perfect.
– user418360
May 22 '17 at 20:06
Can you explain the bounds?
– ifly6
Dec 5 at 16:11
But I need the cylinder surface area, this would be the surface sphere area
– user418360
May 22 '17 at 19:41
But I need the cylinder surface area, this would be the surface sphere area
– user418360
May 22 '17 at 19:41
That is better...
– Doug M
May 22 '17 at 20:00
That is better...
– Doug M
May 22 '17 at 20:00
Thank you so much! it's perfect.
– user418360
May 22 '17 at 20:06
Thank you so much! it's perfect.
– user418360
May 22 '17 at 20:06
Can you explain the bounds?
– ifly6
Dec 5 at 16:11
Can you explain the bounds?
– ifly6
Dec 5 at 16:11
add a comment |
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By area do you mean surface area?
– theyaoster
May 22 '17 at 19:30
yes, I do! surface area
– user418360
May 22 '17 at 19:31