Surface area of a sphere over a discCompute the surface area of an oblate paraboloidFinding the sphere...
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Surface area of a sphere over a disc
Compute the surface area of an oblate paraboloidFinding the sphere surface area using the divergence theorem and sphere volumeWhy does the integral of the surface area of a sphere equal its volume?Calculate surface area of a sphere using the surface integralCalculating surface areaLine integrals - Surface areaSurface integral of function over intersection between plane and unit sphereIntegration over sphere area elementsHow to prove that $text{area}( S_r(x) cap B_R(0)) leq text{area}(S_R(0))$?Volume between cone and sphere of radius $sqrt2$ with surface integral
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What's the surface area of the sphere $x^2 + y^2 + z^2 = 1$ over the disc $(x-1/2)^2 + y^2 le 1/4$ ?
I've tried something, but I don't think it's right, as it's not a "nice answer"
So here is what I've done:
Firstly I parameterized using $x = r cos(t)$, $y= r sin(t)$, $z=z$.
$$dS = sqrt{1+(dz/dx)^2+(dz/dy)^2} dA$$
$$dz/dx=2x , dz/dy=2y $$
Which gives me $$ sqrt{1+4x^2+4y^2} dA = sqrt{1+4r^2} r dr dt$$
Integral then becomes:
S = $int_0^{2pi} int_0^{1/2}sqrt{1+4r^2} r dr dt$
Solving this gives me the result $=0,96$, however I don't think it's right
multivariable-calculus area applications multiple-integral
asked yesterday
LostCoderLostCoder
115
$endgroup$
add a comment |
$begingroup$
What's the surface area of the sphere $x^2 + y^2 + z^2 = 1$ over the disc $(x-1/2)^2 + y^2 le 1/4$ ?
I've tried something, but I don't think it's right, as it's not a "nice answer"
So here is what I've done:
Firstly I parameterized using $x = r cos(t)$, $y= r sin(t)$, $z=z$.
$$dS = sqrt{1+(dz/dx)^2+(dz/dy)^2} dA$$
$$dz/dx=2x , dz/dy=2y $$
Which gives me $$ sqrt{1+4x^2+4y^2} dA = sqrt{1+4r^2} r dr dt$$
Integral then becomes:
S = $int_0^{2pi} int_0^{1/2}sqrt{1+4r^2} r dr dt$
Solving this gives me the result $=0,96$, however I don't think it's right
multivariable-calculus area applications multiple-integral
asked yesterday
LostCoderLostCoder
115
$endgroup$
$begingroup$
Please use MathJax to format posts on this site.
$endgroup$
– saulspatz
yesterday
$begingroup$
I am sorry, last time it made changes by itself, will try to sort it out asap
$endgroup$
– LostCoder
yesterday
$begingroup$
I made some edits to get you started
$endgroup$
– saulspatz
yesterday
$begingroup$
Thank you very much! :)
$endgroup$
– LostCoder
yesterday
$begingroup$
You are integrating over the disk $x^2y^2leqfrac14$ which is not the region specified in the problem
$endgroup$
– saulspatz
yesterday
add a comment |
$begingroup$
What's the surface area of the sphere $x^2 + y^2 + z^2 = 1$ over the disc $(x-1/2)^2 + y^2 le 1/4$ ?
I've tried something, but I don't think it's right, as it's not a "nice answer"
So here is what I've done:
Firstly I parameterized using $x = r cos(t)$, $y= r sin(t)$, $z=z$.
$$dS = sqrt{1+(dz/dx)^2+(dz/dy)^2} dA$$
$$dz/dx=2x , dz/dy=2y $$
Which gives me $$ sqrt{1+4x^2+4y^2} dA = sqrt{1+4r^2} r dr dt$$
Integral then becomes:
S = $int_0^{2pi} int_0^{1/2}sqrt{1+4r^2} r dr dt$
Solving this gives me the result $=0,96$, however I don't think it's right
multivariable-calculus area applications multiple-integral
asked yesterday
LostCoderLostCoder
115
$endgroup$
What's the surface area of the sphere $x^2 + y^2 + z^2 = 1$ over the disc $(x-1/2)^2 + y^2 le 1/4$ ?
I've tried something, but I don't think it's right, as it's not a "nice answer"
So here is what I've done:
Firstly I parameterized using $x = r cos(t)$, $y= r sin(t)$, $z=z$.
$$dS = sqrt{1+(dz/dx)^2+(dz/dy)^2} dA$$
$$dz/dx=2x , dz/dy=2y $$
Which gives me $$ sqrt{1+4x^2+4y^2} dA = sqrt{1+4r^2} r dr dt$$
Integral then becomes:
S = $int_0^{2pi} int_0^{1/2}sqrt{1+4r^2} r dr dt$
Solving this gives me the result $=0,96$, however I don't think it's right
multivariable-calculus area applications multiple-integral
multivariable-calculus area applications multiple-integral
asked yesterday
LostCoderLostCoder
115
asked yesterday
LostCoderLostCoder
115
edited yesterday
LostCoder
asked yesterday
LostCoderLostCoder
115
asked yesterday
LostCoderLostCoder
115
asked yesterday
LostCoderLostCoder
115
115
$begingroup$
Please use MathJax to format posts on this site.
$endgroup$
– saulspatz
yesterday
$begingroup$
I am sorry, last time it made changes by itself, will try to sort it out asap
$endgroup$
– LostCoder
yesterday
$begingroup$
I made some edits to get you started
$endgroup$
– saulspatz
yesterday
$begingroup$
Thank you very much! :)
$endgroup$
– LostCoder
yesterday
$begingroup$
You are integrating over the disk $x^2y^2leqfrac14$ which is not the region specified in the problem
$endgroup$
– saulspatz
yesterday
add a comment |
$begingroup$
Please use MathJax to format posts on this site.
$endgroup$
– saulspatz
yesterday
$begingroup$
I am sorry, last time it made changes by itself, will try to sort it out asap
$endgroup$
– LostCoder
yesterday
$begingroup$
I made some edits to get you started
$endgroup$
– saulspatz
yesterday
$begingroup$
Thank you very much! :)
$endgroup$
– LostCoder
yesterday
$begingroup$
You are integrating over the disk $x^2y^2leqfrac14$ which is not the region specified in the problem
$endgroup$
– saulspatz
yesterday
$begingroup$
Please use MathJax to format posts on this site.
$endgroup$
– saulspatz
yesterday
$begingroup$
Please use MathJax to format posts on this site.
$endgroup$
– saulspatz
yesterday
$begingroup$
I am sorry, last time it made changes by itself, will try to sort it out asap
$endgroup$
– LostCoder
yesterday
$begingroup$
I am sorry, last time it made changes by itself, will try to sort it out asap
$endgroup$
– LostCoder
yesterday
$begingroup$
I made some edits to get you started
$endgroup$
– saulspatz
yesterday
$begingroup$
I made some edits to get you started
$endgroup$
– saulspatz
yesterday
$begingroup$
Thank you very much! :)
$endgroup$
– LostCoder
yesterday
$begingroup$
Thank you very much! :)
$endgroup$
– LostCoder
yesterday
$begingroup$
You are integrating over the disk $x^2y^2leqfrac14$ which is not the region specified in the problem
$endgroup$
– saulspatz
yesterday
$begingroup$
You are integrating over the disk $x^2y^2leqfrac14$ which is not the region specified in the problem
$endgroup$
– saulspatz
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The border of the region in polar coordinates is:
$$(rcostheta - 1/2)^2 + (rsintheta)^2 = 1/4,$$
$$r = costheta,$$
with $thetaincdots$ (Draw the circle to see the interval of variation of $theta$).
answered 19 hours ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
$endgroup$
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
$begingroup$
The border of the region in polar coordinates is:
$$(rcostheta - 1/2)^2 + (rsintheta)^2 = 1/4,$$
$$r = costheta,$$
with $thetaincdots$ (Draw the circle to see the interval of variation of $theta$).
answered 19 hours ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
$endgroup$
add a comment |
$begingroup$
The border of the region in polar coordinates is:
$$(rcostheta - 1/2)^2 + (rsintheta)^2 = 1/4,$$
$$r = costheta,$$
with $thetaincdots$ (Draw the circle to see the interval of variation of $theta$).
answered 19 hours ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
$endgroup$
add a comment |
$begingroup$
The border of the region in polar coordinates is:
$$(rcostheta - 1/2)^2 + (rsintheta)^2 = 1/4,$$
$$r = costheta,$$
with $thetaincdots$ (Draw the circle to see the interval of variation of $theta$).
answered 19 hours ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
$endgroup$
The border of the region in polar coordinates is:
$$(rcostheta - 1/2)^2 + (rsintheta)^2 = 1/4,$$
$$r = costheta,$$
with $thetaincdots$ (Draw the circle to see the interval of variation of $theta$).
answered 19 hours ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
answered 19 hours ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
answered 19 hours ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
answered 19 hours ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
34.5k42971
add a comment |
add a comment |
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$begingroup$
Please use MathJax to format posts on this site.
$endgroup$
– saulspatz
yesterday
$begingroup$
I am sorry, last time it made changes by itself, will try to sort it out asap
$endgroup$
– LostCoder
yesterday
$begingroup$
I made some edits to get you started
$endgroup$
– saulspatz
yesterday
$begingroup$
Thank you very much! :)
$endgroup$
– LostCoder
yesterday
$begingroup$
You are integrating over the disk $x^2y^2leqfrac14$ which is not the region specified in the problem
$endgroup$
– saulspatz
yesterday