Evaluate the line integral $int_C xy^4 ds $ of a half circle












0












$begingroup$



Evaluate the line integral where $C$ is the given curve:
$int_C xy^4 ds $, $C$ it the right half of the circle $x^2+y^2=16$




enter image description here



I was following a similar example in my book and parameterized $t$ by making $x = cos t$ and $y = sin t$ and using the line integral equation, but I got a small fraction while the answer is a large number. I looked at the example twice and checked my problem twice, so I'm not sure what I'm missing. If anyone could help it would be greatly appreciated..






calculus integration multivariable-calculus parametrization







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  • 2




    $begingroup$
    the correct parameterization is x=4*cos(t), y=4*sin(t)
    $endgroup$
    – gimusi
    Nov 26 '17 at 18:39










  • $begingroup$
    I actually was looking at my problem again and just realized that lol
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:40










  • $begingroup$
    alright I plugged in the right parametric values and my radical came out to be 1/4 and the whole thing came out to be 512/5 which is 102.4.. but the right answer is way bigger...???
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:43










  • $begingroup$
    you have to put also a $rho$ to take into account the variables change
    $endgroup$
    – gimusi
    Nov 26 '17 at 18:45






  • 1




    $begingroup$
    If you want someone to point out where you’re going wrong, you should include your work in the question instead of simply writing that you keep getting the wrong answer.
    $endgroup$
    – amd
    Nov 26 '17 at 23:27
















0












$begingroup$



Evaluate the line integral where $C$ is the given curve:
$int_C xy^4 ds $, $C$ it the right half of the circle $x^2+y^2=16$




enter image description here



I was following a similar example in my book and parameterized $t$ by making $x = cos t$ and $y = sin t$ and using the line integral equation, but I got a small fraction while the answer is a large number. I looked at the example twice and checked my problem twice, so I'm not sure what I'm missing. If anyone could help it would be greatly appreciated..






calculus integration multivariable-calculus parametrization







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    the correct parameterization is x=4*cos(t), y=4*sin(t)
    $endgroup$
    – gimusi
    Nov 26 '17 at 18:39










  • $begingroup$
    I actually was looking at my problem again and just realized that lol
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:40










  • $begingroup$
    alright I plugged in the right parametric values and my radical came out to be 1/4 and the whole thing came out to be 512/5 which is 102.4.. but the right answer is way bigger...???
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:43










  • $begingroup$
    you have to put also a $rho$ to take into account the variables change
    $endgroup$
    – gimusi
    Nov 26 '17 at 18:45






  • 1




    $begingroup$
    If you want someone to point out where you’re going wrong, you should include your work in the question instead of simply writing that you keep getting the wrong answer.
    $endgroup$
    – amd
    Nov 26 '17 at 23:27














0












0








0





$begingroup$



Evaluate the line integral where $C$ is the given curve:
$int_C xy^4 ds $, $C$ it the right half of the circle $x^2+y^2=16$




enter image description here



I was following a similar example in my book and parameterized $t$ by making $x = cos t$ and $y = sin t$ and using the line integral equation, but I got a small fraction while the answer is a large number. I looked at the example twice and checked my problem twice, so I'm not sure what I'm missing. If anyone could help it would be greatly appreciated..






calculus integration multivariable-calculus parametrization







share|cite|improve this question











$endgroup$





Evaluate the line integral where $C$ is the given curve:
$int_C xy^4 ds $, $C$ it the right half of the circle $x^2+y^2=16$




enter image description here



I was following a similar example in my book and parameterized $t$ by making $x = cos t$ and $y = sin t$ and using the line integral equation, but I got a small fraction while the answer is a large number. I looked at the example twice and checked my problem twice, so I'm not sure what I'm missing. If anyone could help it would be greatly appreciated..






calculus integration multivariable-calculus parametrization




calculus integration multivariable-calculus parametrization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 2:12









Martin Sleziak

44.7k10119272




44.7k10119272










asked Nov 26 '17 at 18:35









23163546542316354654

306213




306213








  • 2




    $begingroup$
    the correct parameterization is x=4*cos(t), y=4*sin(t)
    $endgroup$
    – gimusi
    Nov 26 '17 at 18:39










  • $begingroup$
    I actually was looking at my problem again and just realized that lol
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:40










  • $begingroup$
    alright I plugged in the right parametric values and my radical came out to be 1/4 and the whole thing came out to be 512/5 which is 102.4.. but the right answer is way bigger...???
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:43










  • $begingroup$
    you have to put also a $rho$ to take into account the variables change
    $endgroup$
    – gimusi
    Nov 26 '17 at 18:45






  • 1




    $begingroup$
    If you want someone to point out where you’re going wrong, you should include your work in the question instead of simply writing that you keep getting the wrong answer.
    $endgroup$
    – amd
    Nov 26 '17 at 23:27














  • 2




    $begingroup$
    the correct parameterization is x=4*cos(t), y=4*sin(t)
    $endgroup$
    – gimusi
    Nov 26 '17 at 18:39










  • $begingroup$
    I actually was looking at my problem again and just realized that lol
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:40










  • $begingroup$
    alright I plugged in the right parametric values and my radical came out to be 1/4 and the whole thing came out to be 512/5 which is 102.4.. but the right answer is way bigger...???
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:43










  • $begingroup$
    you have to put also a $rho$ to take into account the variables change
    $endgroup$
    – gimusi
    Nov 26 '17 at 18:45






  • 1




    $begingroup$
    If you want someone to point out where you’re going wrong, you should include your work in the question instead of simply writing that you keep getting the wrong answer.
    $endgroup$
    – amd
    Nov 26 '17 at 23:27








2




2




$begingroup$
the correct parameterization is x=4*cos(t), y=4*sin(t)
$endgroup$
– gimusi
Nov 26 '17 at 18:39




$begingroup$
the correct parameterization is x=4*cos(t), y=4*sin(t)
$endgroup$
– gimusi
Nov 26 '17 at 18:39












$begingroup$
I actually was looking at my problem again and just realized that lol
$endgroup$
– 2316354654
Nov 26 '17 at 18:40




$begingroup$
I actually was looking at my problem again and just realized that lol
$endgroup$
– 2316354654
Nov 26 '17 at 18:40












$begingroup$
alright I plugged in the right parametric values and my radical came out to be 1/4 and the whole thing came out to be 512/5 which is 102.4.. but the right answer is way bigger...???
$endgroup$
– 2316354654
Nov 26 '17 at 18:43




$begingroup$
alright I plugged in the right parametric values and my radical came out to be 1/4 and the whole thing came out to be 512/5 which is 102.4.. but the right answer is way bigger...???
$endgroup$
– 2316354654
Nov 26 '17 at 18:43












$begingroup$
you have to put also a $rho$ to take into account the variables change
$endgroup$
– gimusi
Nov 26 '17 at 18:45




$begingroup$
you have to put also a $rho$ to take into account the variables change
$endgroup$
– gimusi
Nov 26 '17 at 18:45




1




1




$begingroup$
If you want someone to point out where you’re going wrong, you should include your work in the question instead of simply writing that you keep getting the wrong answer.
$endgroup$
– amd
Nov 26 '17 at 23:27




$begingroup$
If you want someone to point out where you’re going wrong, you should include your work in the question instead of simply writing that you keep getting the wrong answer.
$endgroup$
– amd
Nov 26 '17 at 23:27










3 Answers
3






active

oldest

votes


















0












$begingroup$

$$r(t)=(4cos t,,4sin t);,;;-fracpi2le tlefracpi2implies text{our integral is}$$



$$int_{-fracpi2}^{fracpi2}4cos t(4sin t)^4left|r'(t)right|dt=1,024int_{-fracpi2}^{fracpi2}cos t,sin^4tcdot4,dt=left.frac{4,096}5sin^5tright|_{-fracpi2}^{fracpi2}=frac{8,192}5$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    how did you get the 4 for the norm of r'(t) ?
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:44










  • $begingroup$
    why do you use commata?
    $endgroup$
    – klirk
    Nov 26 '17 at 18:47










  • $begingroup$
    wait nevermind i see it.
    $endgroup$
    – 2316354654
    Nov 26 '17 at 18:47



















1












$begingroup$

You forgot to include the radius. The substitution $x= cos t$, $y= sin t$ parametrizes a circle of radius $1$.

But $C$ is a subset of a circle of radius $4$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The correct parameterization is $x=4cos(t), y=4sin(t)$.






    share|cite|improve this answer











    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      $$r(t)=(4cos t,,4sin t);,;;-fracpi2le tlefracpi2implies text{our integral is}$$



      $$int_{-fracpi2}^{fracpi2}4cos t(4sin t)^4left|r'(t)right|dt=1,024int_{-fracpi2}^{fracpi2}cos t,sin^4tcdot4,dt=left.frac{4,096}5sin^5tright|_{-fracpi2}^{fracpi2}=frac{8,192}5$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        how did you get the 4 for the norm of r'(t) ?
        $endgroup$
        – 2316354654
        Nov 26 '17 at 18:44










      • $begingroup$
        why do you use commata?
        $endgroup$
        – klirk
        Nov 26 '17 at 18:47










      • $begingroup$
        wait nevermind i see it.
        $endgroup$
        – 2316354654
        Nov 26 '17 at 18:47
















      0












      $begingroup$

      $$r(t)=(4cos t,,4sin t);,;;-fracpi2le tlefracpi2implies text{our integral is}$$



      $$int_{-fracpi2}^{fracpi2}4cos t(4sin t)^4left|r'(t)right|dt=1,024int_{-fracpi2}^{fracpi2}cos t,sin^4tcdot4,dt=left.frac{4,096}5sin^5tright|_{-fracpi2}^{fracpi2}=frac{8,192}5$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        how did you get the 4 for the norm of r'(t) ?
        $endgroup$
        – 2316354654
        Nov 26 '17 at 18:44










      • $begingroup$
        why do you use commata?
        $endgroup$
        – klirk
        Nov 26 '17 at 18:47










      • $begingroup$
        wait nevermind i see it.
        $endgroup$
        – 2316354654
        Nov 26 '17 at 18:47














      0












      0








      0





      $begingroup$

      $$r(t)=(4cos t,,4sin t);,;;-fracpi2le tlefracpi2implies text{our integral is}$$



      $$int_{-fracpi2}^{fracpi2}4cos t(4sin t)^4left|r'(t)right|dt=1,024int_{-fracpi2}^{fracpi2}cos t,sin^4tcdot4,dt=left.frac{4,096}5sin^5tright|_{-fracpi2}^{fracpi2}=frac{8,192}5$$






      share|cite|improve this answer









      $endgroup$



      $$r(t)=(4cos t,,4sin t);,;;-fracpi2le tlefracpi2implies text{our integral is}$$



      $$int_{-fracpi2}^{fracpi2}4cos t(4sin t)^4left|r'(t)right|dt=1,024int_{-fracpi2}^{fracpi2}cos t,sin^4tcdot4,dt=left.frac{4,096}5sin^5tright|_{-fracpi2}^{fracpi2}=frac{8,192}5$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 26 '17 at 18:42









      DonAntonioDonAntonio

      179k1494230




      179k1494230












      • $begingroup$
        how did you get the 4 for the norm of r'(t) ?
        $endgroup$
        – 2316354654
        Nov 26 '17 at 18:44










      • $begingroup$
        why do you use commata?
        $endgroup$
        – klirk
        Nov 26 '17 at 18:47










      • $begingroup$
        wait nevermind i see it.
        $endgroup$
        – 2316354654
        Nov 26 '17 at 18:47


















      • $begingroup$
        how did you get the 4 for the norm of r'(t) ?
        $endgroup$
        – 2316354654
        Nov 26 '17 at 18:44










      • $begingroup$
        why do you use commata?
        $endgroup$
        – klirk
        Nov 26 '17 at 18:47










      • $begingroup$
        wait nevermind i see it.
        $endgroup$
        – 2316354654
        Nov 26 '17 at 18:47
















      $begingroup$
      how did you get the 4 for the norm of r'(t) ?
      $endgroup$
      – 2316354654
      Nov 26 '17 at 18:44




      $begingroup$
      how did you get the 4 for the norm of r'(t) ?
      $endgroup$
      – 2316354654
      Nov 26 '17 at 18:44












      $begingroup$
      why do you use commata?
      $endgroup$
      – klirk
      Nov 26 '17 at 18:47




      $begingroup$
      why do you use commata?
      $endgroup$
      – klirk
      Nov 26 '17 at 18:47












      $begingroup$
      wait nevermind i see it.
      $endgroup$
      – 2316354654
      Nov 26 '17 at 18:47




      $begingroup$
      wait nevermind i see it.
      $endgroup$
      – 2316354654
      Nov 26 '17 at 18:47











      1












      $begingroup$

      You forgot to include the radius. The substitution $x= cos t$, $y= sin t$ parametrizes a circle of radius $1$.

      But $C$ is a subset of a circle of radius $4$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        You forgot to include the radius. The substitution $x= cos t$, $y= sin t$ parametrizes a circle of radius $1$.

        But $C$ is a subset of a circle of radius $4$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You forgot to include the radius. The substitution $x= cos t$, $y= sin t$ parametrizes a circle of radius $1$.

          But $C$ is a subset of a circle of radius $4$.






          share|cite|improve this answer









          $endgroup$



          You forgot to include the radius. The substitution $x= cos t$, $y= sin t$ parametrizes a circle of radius $1$.

          But $C$ is a subset of a circle of radius $4$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '17 at 18:40









          klirkklirk

          2,788630




          2,788630























              0












              $begingroup$

              The correct parameterization is $x=4cos(t), y=4sin(t)$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                The correct parameterization is $x=4cos(t), y=4sin(t)$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The correct parameterization is $x=4cos(t), y=4sin(t)$.






                  share|cite|improve this answer











                  $endgroup$



                  The correct parameterization is $x=4cos(t), y=4sin(t)$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 11 '18 at 14:19









                  Davide Giraudo

                  127k16151264




                  127k16151264










                  answered Nov 26 '17 at 18:40









                  gimusigimusi

                  92.9k84494




                  92.9k84494






























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