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Find The Area Of $gamma(t)=(at-asin t,a-acos t)$
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Find the area of $gamma(t)=(at-asin t,a-acos t)$ where $tin[0,2pi], a>0$
I have bought to use Green's theorem $A=frac12oint_gamma(-ydx+xdy)$
$gamma(t)=(at-asin t,a-acos t)$
$gamma'(t)=(a-acos t,asin t)$
$A=frac12oint_gamma(-ydx+xdy)=frac12int_0^2pi(-(a-acos t)^2+a^2tsin t-a^2sin^2 t )dt=\=frac12int_0^2pi(-2a^2+2a^2cos t+a^2sin t)dt=frac12(-2a^2t+2a^2sin t-a^2cos t)|_0^2pi=-4a^2pi$
Which can not be as it is negative, where did it went wrong? I did not use the fact that $a>0$ maybe I had to try other integral
multivariable-calculus
asked Jul 31 at 14:08
newhere
742310
add a comment |Â
up vote
-1
down vote
favorite
Find the area of $gamma(t)=(at-asin t,a-acos t)$ where $tin[0,2pi], a>0$
I have bought to use Green's theorem $A=frac12oint_gamma(-ydx+xdy)$
$gamma(t)=(at-asin t,a-acos t)$
$gamma'(t)=(a-acos t,asin t)$
$A=frac12oint_gamma(-ydx+xdy)=frac12int_0^2pi(-(a-acos t)^2+a^2tsin t-a^2sin^2 t )dt=\=frac12int_0^2pi(-2a^2+2a^2cos t+a^2sin t)dt=frac12(-2a^2t+2a^2sin t-a^2cos t)|_0^2pi=-4a^2pi$
Which can not be as it is negative, where did it went wrong? I did not use the fact that $a>0$ maybe I had to try other integral
multivariable-calculus
asked Jul 31 at 14:08
newhere
742310
1
Your $gamma$ is not closed, don't you need that?
– Kusma
Jul 31 at 14:15
Your $gamma$ should be closed and positively oriented.
– Sobi
Jul 31 at 14:17
I don't think you can find the area of a curve. I think you mean the area enclosed by the curve. Before using Green's Theorem you have to check to see if it is a simple closed curve. Also, maybe your algebra is just sloppy--where did the t sin t term go? Are you just missing a t? Maybe a > 0 has something to do with the way the curve is drawn, like its orientation. You need a positive orientation to apply Green's Theorem.
– Leonard Blackburn
Jul 31 at 14:18
Over $[0,2pi]$ the curve traces out one cycle of a cycloid. This is not a closed curve. If we close it with a line from $(2pi a,0)$ to $(0,0)$, then the curve goes around the region in a clockwise direction, and our equations are oreinted for a counter-clockwise path. The reversed orientation flips the sign.
– Doug M
Jul 31 at 14:35
@DougM so I can integrate from $2pi$ to $0$ or to flip the sign, as for closing the curve, I need to remove the integral from $0$ to $2pi$ there $y=0$?
– newhere
Jul 31 at 14:53
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Find the area of $gamma(t)=(at-asin t,a-acos t)$ where $tin[0,2pi], a>0$
I have bought to use Green's theorem $A=frac12oint_gamma(-ydx+xdy)$
$gamma(t)=(at-asin t,a-acos t)$
$gamma'(t)=(a-acos t,asin t)$
$A=frac12oint_gamma(-ydx+xdy)=frac12int_0^2pi(-(a-acos t)^2+a^2tsin t-a^2sin^2 t )dt=\=frac12int_0^2pi(-2a^2+2a^2cos t+a^2sin t)dt=frac12(-2a^2t+2a^2sin t-a^2cos t)|_0^2pi=-4a^2pi$
Which can not be as it is negative, where did it went wrong? I did not use the fact that $a>0$ maybe I had to try other integral
multivariable-calculus
asked Jul 31 at 14:08
newhere
742310
Find the area of $gamma(t)=(at-asin t,a-acos t)$ where $tin[0,2pi], a>0$
I have bought to use Green's theorem $A=frac12oint_gamma(-ydx+xdy)$
$gamma(t)=(at-asin t,a-acos t)$
$gamma'(t)=(a-acos t,asin t)$
$A=frac12oint_gamma(-ydx+xdy)=frac12int_0^2pi(-(a-acos t)^2+a^2tsin t-a^2sin^2 t )dt=\=frac12int_0^2pi(-2a^2+2a^2cos t+a^2sin t)dt=frac12(-2a^2t+2a^2sin t-a^2cos t)|_0^2pi=-4a^2pi$
Which can not be as it is negative, where did it went wrong? I did not use the fact that $a>0$ maybe I had to try other integral
multivariable-calculus
asked Jul 31 at 14:08
newhere
742310
asked Jul 31 at 14:08
newhere
742310
asked Jul 31 at 14:08
newhere
742310
asked Jul 31 at 14:08
newhere
742310
742310
1
Your $gamma$ is not closed, don't you need that?
– Kusma
Jul 31 at 14:15
Your $gamma$ should be closed and positively oriented.
– Sobi
Jul 31 at 14:17
I don't think you can find the area of a curve. I think you mean the area enclosed by the curve. Before using Green's Theorem you have to check to see if it is a simple closed curve. Also, maybe your algebra is just sloppy--where did the t sin t term go? Are you just missing a t? Maybe a > 0 has something to do with the way the curve is drawn, like its orientation. You need a positive orientation to apply Green's Theorem.
– Leonard Blackburn
Jul 31 at 14:18
Over $[0,2pi]$ the curve traces out one cycle of a cycloid. This is not a closed curve. If we close it with a line from $(2pi a,0)$ to $(0,0)$, then the curve goes around the region in a clockwise direction, and our equations are oreinted for a counter-clockwise path. The reversed orientation flips the sign.
– Doug M
Jul 31 at 14:35
@DougM so I can integrate from $2pi$ to $0$ or to flip the sign, as for closing the curve, I need to remove the integral from $0$ to $2pi$ there $y=0$?
– newhere
Jul 31 at 14:53
add a comment |Â
1
Your $gamma$ is not closed, don't you need that?
– Kusma
Jul 31 at 14:15
Your $gamma$ should be closed and positively oriented.
– Sobi
Jul 31 at 14:17
I don't think you can find the area of a curve. I think you mean the area enclosed by the curve. Before using Green's Theorem you have to check to see if it is a simple closed curve. Also, maybe your algebra is just sloppy--where did the t sin t term go? Are you just missing a t? Maybe a > 0 has something to do with the way the curve is drawn, like its orientation. You need a positive orientation to apply Green's Theorem.
– Leonard Blackburn
Jul 31 at 14:18
Over $[0,2pi]$ the curve traces out one cycle of a cycloid. This is not a closed curve. If we close it with a line from $(2pi a,0)$ to $(0,0)$, then the curve goes around the region in a clockwise direction, and our equations are oreinted for a counter-clockwise path. The reversed orientation flips the sign.
– Doug M
Jul 31 at 14:35
@DougM so I can integrate from $2pi$ to $0$ or to flip the sign, as for closing the curve, I need to remove the integral from $0$ to $2pi$ there $y=0$?
– newhere
Jul 31 at 14:53
1
1
Your $gamma$ is not closed, don't you need that?
– Kusma
Jul 31 at 14:15
Your $gamma$ is not closed, don't you need that?
– Kusma
Jul 31 at 14:15
Your $gamma$ should be closed and positively oriented.
– Sobi
Jul 31 at 14:17
Your $gamma$ should be closed and positively oriented.
– Sobi
Jul 31 at 14:17
I don't think you can find the area of a curve. I think you mean the area enclosed by the curve. Before using Green's Theorem you have to check to see if it is a simple closed curve. Also, maybe your algebra is just sloppy--where did the t sin t term go? Are you just missing a t? Maybe a > 0 has something to do with the way the curve is drawn, like its orientation. You need a positive orientation to apply Green's Theorem.
– Leonard Blackburn
Jul 31 at 14:18
I don't think you can find the area of a curve. I think you mean the area enclosed by the curve. Before using Green's Theorem you have to check to see if it is a simple closed curve. Also, maybe your algebra is just sloppy--where did the t sin t term go? Are you just missing a t? Maybe a > 0 has something to do with the way the curve is drawn, like its orientation. You need a positive orientation to apply Green's Theorem.
– Leonard Blackburn
Jul 31 at 14:18
Over $[0,2pi]$ the curve traces out one cycle of a cycloid. This is not a closed curve. If we close it with a line from $(2pi a,0)$ to $(0,0)$, then the curve goes around the region in a clockwise direction, and our equations are oreinted for a counter-clockwise path. The reversed orientation flips the sign.
– Doug M
Jul 31 at 14:35
Over $[0,2pi]$ the curve traces out one cycle of a cycloid. This is not a closed curve. If we close it with a line from $(2pi a,0)$ to $(0,0)$, then the curve goes around the region in a clockwise direction, and our equations are oreinted for a counter-clockwise path. The reversed orientation flips the sign.
– Doug M
Jul 31 at 14:35
@DougM so I can integrate from $2pi$ to $0$ or to flip the sign, as for closing the curve, I need to remove the integral from $0$ to $2pi$ there $y=0$?
– newhere
Jul 31 at 14:53
@DougM so I can integrate from $2pi$ to $0$ or to flip the sign, as for closing the curve, I need to remove the integral from $0$ to $2pi$ there $y=0$?
– newhere
Jul 31 at 14:53
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
The red line is your curve.
If you don't close the curve, you don't have an enclosed area. The line y=0 will suffice to close the curve.
As for your integration: You can think $frac 12 (x dy - y dx)$ as the green area. And the integral will be the sum of triangles with small changes in $(x,y)$ Since we are traversing clockwise, that area is $frac 12 (x dy - y dx)$ which is of course the negative of the more familiar $frac 12 (y dx - x dy)$ associated with counter-clockwise paths.
As far what does the line y=0 do to your integration? As all triangles connecting the origin to this path are degenerate triangles of 0 area, it has 0 impact.
answered Jul 31 at 15:10
Doug M
39k31749
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The red line is your curve.
If you don't close the curve, you don't have an enclosed area. The line y=0 will suffice to close the curve.
As for your integration: You can think $frac 12 (x dy - y dx)$ as the green area. And the integral will be the sum of triangles with small changes in $(x,y)$ Since we are traversing clockwise, that area is $frac 12 (x dy - y dx)$ which is of course the negative of the more familiar $frac 12 (y dx - x dy)$ associated with counter-clockwise paths.
As far what does the line y=0 do to your integration? As all triangles connecting the origin to this path are degenerate triangles of 0 area, it has 0 impact.
answered Jul 31 at 15:10
Doug M
39k31749
add a comment |Â
up vote
1
down vote
accepted
The red line is your curve.
If you don't close the curve, you don't have an enclosed area. The line y=0 will suffice to close the curve.
As for your integration: You can think $frac 12 (x dy - y dx)$ as the green area. And the integral will be the sum of triangles with small changes in $(x,y)$ Since we are traversing clockwise, that area is $frac 12 (x dy - y dx)$ which is of course the negative of the more familiar $frac 12 (y dx - x dy)$ associated with counter-clockwise paths.
As far what does the line y=0 do to your integration? As all triangles connecting the origin to this path are degenerate triangles of 0 area, it has 0 impact.
answered Jul 31 at 15:10
Doug M
39k31749
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The red line is your curve.
If you don't close the curve, you don't have an enclosed area. The line y=0 will suffice to close the curve.
As for your integration: You can think $frac 12 (x dy - y dx)$ as the green area. And the integral will be the sum of triangles with small changes in $(x,y)$ Since we are traversing clockwise, that area is $frac 12 (x dy - y dx)$ which is of course the negative of the more familiar $frac 12 (y dx - x dy)$ associated with counter-clockwise paths.
As far what does the line y=0 do to your integration? As all triangles connecting the origin to this path are degenerate triangles of 0 area, it has 0 impact.
answered Jul 31 at 15:10
Doug M
39k31749
The red line is your curve.
If you don't close the curve, you don't have an enclosed area. The line y=0 will suffice to close the curve.
As for your integration: You can think $frac 12 (x dy - y dx)$ as the green area. And the integral will be the sum of triangles with small changes in $(x,y)$ Since we are traversing clockwise, that area is $frac 12 (x dy - y dx)$ which is of course the negative of the more familiar $frac 12 (y dx - x dy)$ associated with counter-clockwise paths.
As far what does the line y=0 do to your integration? As all triangles connecting the origin to this path are degenerate triangles of 0 area, it has 0 impact.
answered Jul 31 at 15:10
Doug M
39k31749
answered Jul 31 at 15:10
Doug M
39k31749
answered Jul 31 at 15:10
Doug M
39k31749
answered Jul 31 at 15:10
Doug M
39k31749
39k31749
add a comment |Â
add a comment |Â
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1
Your $gamma$ is not closed, don't you need that?
– Kusma
Jul 31 at 14:15
Your $gamma$ should be closed and positively oriented.
– Sobi
Jul 31 at 14:17
I don't think you can find the area of a curve. I think you mean the area enclosed by the curve. Before using Green's Theorem you have to check to see if it is a simple closed curve. Also, maybe your algebra is just sloppy--where did the t sin t term go? Are you just missing a t? Maybe a > 0 has something to do with the way the curve is drawn, like its orientation. You need a positive orientation to apply Green's Theorem.
– Leonard Blackburn
Jul 31 at 14:18
Over $[0,2pi]$ the curve traces out one cycle of a cycloid. This is not a closed curve. If we close it with a line from $(2pi a,0)$ to $(0,0)$, then the curve goes around the region in a clockwise direction, and our equations are oreinted for a counter-clockwise path. The reversed orientation flips the sign.
– Doug M
Jul 31 at 14:35
@DougM so I can integrate from $2pi$ to $0$ or to flip the sign, as for closing the curve, I need to remove the integral from $0$ to $2pi$ there $y=0$?
– newhere
Jul 31 at 14:53