Mixing
calculate this line integral
$begingroup$
this is my curve :
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{t})$
$||r'(t)||=sqrt{sin^2{t}+1}$
I have to calculate: $int{(y+1)}ds$
So I have : $int_{0}^{3pi}sin tsqrt{sin^2{t}+1}dt$
Is this right? If it is, can you help me figure out how to compute this integral? I tried a lot of substitution but I can't get it any simpler.
integration indefinite-integrals line-integrals
edited Jan 19 at 23:52
David G. Stork
11k41432
asked Jan 19 at 23:31
NPLSNPLS
7612
$endgroup$
add a comment |
$begingroup$
this is my curve :
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{t})$
$||r'(t)||=sqrt{sin^2{t}+1}$
I have to calculate: $int{(y+1)}ds$
So I have : $int_{0}^{3pi}sin tsqrt{sin^2{t}+1}dt$
Is this right? If it is, can you help me figure out how to compute this integral? I tried a lot of substitution but I can't get it any simpler.
integration indefinite-integrals line-integrals
edited Jan 19 at 23:52
David G. Stork
11k41432
asked Jan 19 at 23:31
NPLSNPLS
7612
$endgroup$
2
$begingroup$
You have a mistake in differentiation of $cosfrac{t}{2}$ (you skipped the 1/2 in $sin$).
$endgroup$
– orion
Jan 19 at 23:43
$begingroup$
@orion it’s $2cosfrac{t}{2}$, so he doesn’t.
$endgroup$
– cluelessatthis
Jan 20 at 0:45
$begingroup$
It should be $-sin frac{t}{2}$, not $-sin t$. The half doesn't disappear from inside the trigonometric function after differentiation.
$endgroup$
– orion
Jan 20 at 0:52
add a comment |
$begingroup$
this is my curve :
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{t})$
$||r'(t)||=sqrt{sin^2{t}+1}$
I have to calculate: $int{(y+1)}ds$
So I have : $int_{0}^{3pi}sin tsqrt{sin^2{t}+1}dt$
Is this right? If it is, can you help me figure out how to compute this integral? I tried a lot of substitution but I can't get it any simpler.
integration indefinite-integrals line-integrals
edited Jan 19 at 23:52
David G. Stork
11k41432
asked Jan 19 at 23:31
NPLSNPLS
7612
$endgroup$
this is my curve :
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{t})$
$||r'(t)||=sqrt{sin^2{t}+1}$
I have to calculate: $int{(y+1)}ds$
So I have : $int_{0}^{3pi}sin tsqrt{sin^2{t}+1}dt$
Is this right? If it is, can you help me figure out how to compute this integral? I tried a lot of substitution but I can't get it any simpler.
integration indefinite-integrals line-integrals
integration indefinite-integrals line-integrals
edited Jan 19 at 23:52
David G. Stork
11k41432
asked Jan 19 at 23:31
NPLSNPLS
7612
edited Jan 19 at 23:52
David G. Stork
11k41432
asked Jan 19 at 23:31
NPLSNPLS
7612
edited Jan 19 at 23:52
David G. Stork
11k41432
edited Jan 19 at 23:52
David G. Stork
11k41432
edited Jan 19 at 23:52
David G. Stork
11k41432
11k41432
asked Jan 19 at 23:31
NPLSNPLS
7612
asked Jan 19 at 23:31
NPLSNPLS
7612
asked Jan 19 at 23:31
NPLSNPLS
7612
7612
2
$begingroup$
You have a mistake in differentiation of $cosfrac{t}{2}$ (you skipped the 1/2 in $sin$).
$endgroup$
– orion
Jan 19 at 23:43
$begingroup$
@orion it’s $2cosfrac{t}{2}$, so he doesn’t.
$endgroup$
– cluelessatthis
Jan 20 at 0:45
$begingroup$
It should be $-sin frac{t}{2}$, not $-sin t$. The half doesn't disappear from inside the trigonometric function after differentiation.
$endgroup$
– orion
Jan 20 at 0:52
add a comment |
2
$begingroup$
You have a mistake in differentiation of $cosfrac{t}{2}$ (you skipped the 1/2 in $sin$).
$endgroup$
– orion
Jan 19 at 23:43
$begingroup$
@orion it’s $2cosfrac{t}{2}$, so he doesn’t.
$endgroup$
– cluelessatthis
Jan 20 at 0:45
$begingroup$
It should be $-sin frac{t}{2}$, not $-sin t$. The half doesn't disappear from inside the trigonometric function after differentiation.
$endgroup$
– orion
Jan 20 at 0:52
2
2
$begingroup$
You have a mistake in differentiation of $cosfrac{t}{2}$ (you skipped the 1/2 in $sin$).
$endgroup$
– orion
Jan 19 at 23:43
$begingroup$
You have a mistake in differentiation of $cosfrac{t}{2}$ (you skipped the 1/2 in $sin$).
$endgroup$
– orion
Jan 19 at 23:43
$begingroup$
@orion it’s $2cosfrac{t}{2}$, so he doesn’t.
$endgroup$
– cluelessatthis
Jan 20 at 0:45
$begingroup$
@orion it’s $2cosfrac{t}{2}$, so he doesn’t.
$endgroup$
– cluelessatthis
Jan 20 at 0:45
$begingroup$
It should be $-sin frac{t}{2}$, not $-sin t$. The half doesn't disappear from inside the trigonometric function after differentiation.
$endgroup$
– orion
Jan 20 at 0:52
$begingroup$
It should be $-sin frac{t}{2}$, not $-sin t$. The half doesn't disappear from inside the trigonometric function after differentiation.
$endgroup$
– orion
Jan 20 at 0:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{frac{t}{2}})$ <- You forgot sin(t/2)
$||r'(t)||=sqrt{sin^2{frac{t}{2}}+1}$
Calculate: $int{(y+1)}ds$
-> $I = int_{0}^{3pi}sin tsqrt{sin^2{frac{t}{2}}+1}dt$
Apply subsitution $u = sin^2(frac{t}{2}) + 1$
$du = frac{sin{t}}{2} dt$
$t = 0, u = 1. t = 3pi,u = 2.$
$I = 2int_{1}^{2}sqrt{u}textrm{ }du = frac{4}{3}(2^frac{3}{2}-1^frac{3}{2})=frac{4}{3}times(2sqrt{2}-1)=frac{8sqrt{2}-4}{3}$
answered Jan 20 at 2:21
Gareth MaGareth Ma
526215
$endgroup$
$begingroup$
why is It $du=frac{sint}{2}$ ? how did you get $frac{sint}{2}$?
$endgroup$
– NPLS
Jan 20 at 8:44
$begingroup$
$u = sin^2(frac{t}{2}); du = 2sin(frac{t}{2})cos(frac{t}{2})times frac{1}{2}$ By Chain Rule. Simplify.
$endgroup$
– Gareth Ma
Jan 21 at 9:05
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{frac{t}{2}})$ <- You forgot sin(t/2)
$||r'(t)||=sqrt{sin^2{frac{t}{2}}+1}$
Calculate: $int{(y+1)}ds$
-> $I = int_{0}^{3pi}sin tsqrt{sin^2{frac{t}{2}}+1}dt$
Apply subsitution $u = sin^2(frac{t}{2}) + 1$
$du = frac{sin{t}}{2} dt$
$t = 0, u = 1. t = 3pi,u = 2.$
$I = 2int_{1}^{2}sqrt{u}textrm{ }du = frac{4}{3}(2^frac{3}{2}-1^frac{3}{2})=frac{4}{3}times(2sqrt{2}-1)=frac{8sqrt{2}-4}{3}$
answered Jan 20 at 2:21
Gareth MaGareth Ma
526215
$endgroup$
$begingroup$
why is It $du=frac{sint}{2}$ ? how did you get $frac{sint}{2}$?
$endgroup$
– NPLS
Jan 20 at 8:44
$begingroup$
$u = sin^2(frac{t}{2}); du = 2sin(frac{t}{2})cos(frac{t}{2})times frac{1}{2}$ By Chain Rule. Simplify.
$endgroup$
– Gareth Ma
Jan 21 at 9:05
add a comment |
$begingroup$
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{frac{t}{2}})$ <- You forgot sin(t/2)
$||r'(t)||=sqrt{sin^2{frac{t}{2}}+1}$
Calculate: $int{(y+1)}ds$
-> $I = int_{0}^{3pi}sin tsqrt{sin^2{frac{t}{2}}+1}dt$
Apply subsitution $u = sin^2(frac{t}{2}) + 1$
$du = frac{sin{t}}{2} dt$
$t = 0, u = 1. t = 3pi,u = 2.$
$I = 2int_{1}^{2}sqrt{u}textrm{ }du = frac{4}{3}(2^frac{3}{2}-1^frac{3}{2})=frac{4}{3}times(2sqrt{2}-1)=frac{8sqrt{2}-4}{3}$
answered Jan 20 at 2:21
Gareth MaGareth Ma
526215
$endgroup$
$begingroup$
why is It $du=frac{sint}{2}$ ? how did you get $frac{sint}{2}$?
$endgroup$
– NPLS
Jan 20 at 8:44
$begingroup$
$u = sin^2(frac{t}{2}); du = 2sin(frac{t}{2})cos(frac{t}{2})times frac{1}{2}$ By Chain Rule. Simplify.
$endgroup$
– Gareth Ma
Jan 21 at 9:05
add a comment |
$begingroup$
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{frac{t}{2}})$ <- You forgot sin(t/2)
$||r'(t)||=sqrt{sin^2{frac{t}{2}}+1}$
Calculate: $int{(y+1)}ds$
-> $I = int_{0}^{3pi}sin tsqrt{sin^2{frac{t}{2}}+1}dt$
Apply subsitution $u = sin^2(frac{t}{2}) + 1$
$du = frac{sin{t}}{2} dt$
$t = 0, u = 1. t = 3pi,u = 2.$
$I = 2int_{1}^{2}sqrt{u}textrm{ }du = frac{4}{3}(2^frac{3}{2}-1^frac{3}{2})=frac{4}{3}times(2sqrt{2}-1)=frac{8sqrt{2}-4}{3}$
answered Jan 20 at 2:21
Gareth MaGareth Ma
526215
$endgroup$
$r(t)=(cos{t},sin{t}-1,2cos{frac{t}{2}})$ , $t=[0,3pi]$
$r'(t)=(-sin{t},cos{t},-sin{frac{t}{2}})$ <- You forgot sin(t/2)
$||r'(t)||=sqrt{sin^2{frac{t}{2}}+1}$
Calculate: $int{(y+1)}ds$
-> $I = int_{0}^{3pi}sin tsqrt{sin^2{frac{t}{2}}+1}dt$
Apply subsitution $u = sin^2(frac{t}{2}) + 1$
$du = frac{sin{t}}{2} dt$
$t = 0, u = 1. t = 3pi,u = 2.$
$I = 2int_{1}^{2}sqrt{u}textrm{ }du = frac{4}{3}(2^frac{3}{2}-1^frac{3}{2})=frac{4}{3}times(2sqrt{2}-1)=frac{8sqrt{2}-4}{3}$
answered Jan 20 at 2:21
Gareth MaGareth Ma
526215
edited Jan 21 at 9:06
answered Jan 20 at 2:21
Gareth MaGareth Ma
526215
answered Jan 20 at 2:21
Gareth MaGareth Ma
526215
answered Jan 20 at 2:21
Gareth MaGareth Ma
526215
526215
$begingroup$
why is It $du=frac{sint}{2}$ ? how did you get $frac{sint}{2}$?
$endgroup$
– NPLS
Jan 20 at 8:44
$begingroup$
$u = sin^2(frac{t}{2}); du = 2sin(frac{t}{2})cos(frac{t}{2})times frac{1}{2}$ By Chain Rule. Simplify.
$endgroup$
– Gareth Ma
Jan 21 at 9:05
add a comment |
$begingroup$
why is It $du=frac{sint}{2}$ ? how did you get $frac{sint}{2}$?
$endgroup$
– NPLS
Jan 20 at 8:44
$begingroup$
$u = sin^2(frac{t}{2}); du = 2sin(frac{t}{2})cos(frac{t}{2})times frac{1}{2}$ By Chain Rule. Simplify.
$endgroup$
– Gareth Ma
Jan 21 at 9:05
$begingroup$
why is It $du=frac{sint}{2}$ ? how did you get $frac{sint}{2}$?
$endgroup$
– NPLS
Jan 20 at 8:44
$begingroup$
why is It $du=frac{sint}{2}$ ? how did you get $frac{sint}{2}$?
$endgroup$
– NPLS
Jan 20 at 8:44
$begingroup$
$u = sin^2(frac{t}{2}); du = 2sin(frac{t}{2})cos(frac{t}{2})times frac{1}{2}$ By Chain Rule. Simplify.
$endgroup$
– Gareth Ma
Jan 21 at 9:05
$begingroup$
$u = sin^2(frac{t}{2}); du = 2sin(frac{t}{2})cos(frac{t}{2})times frac{1}{2}$ By Chain Rule. Simplify.
$endgroup$
– Gareth Ma
Jan 21 at 9:05
add a comment |
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2
$begingroup$
You have a mistake in differentiation of $cosfrac{t}{2}$ (you skipped the 1/2 in $sin$).
$endgroup$
– orion
Jan 19 at 23:43
$begingroup$
@orion it’s $2cosfrac{t}{2}$, so he doesn’t.
$endgroup$
– cluelessatthis
Jan 20 at 0:45
$begingroup$
It should be $-sin frac{t}{2}$, not $-sin t$. The half doesn't disappear from inside the trigonometric function after differentiation.
$endgroup$
– orion
Jan 20 at 0:52