Evaluate the line integral integral_c xy^4 ds, where C is the right half of the

Question

Evaluate the line integral integral_c xy^4 ds, where C is the right half of the circle x^2 + y^2 = 16. Evaluate the line integral integral_c.y^2zds, where C is the line segment from (3, 1, 2) to (1, 2, 5). Evaluate the line integral integral_c F dr. where F(x, y) = (xy^2, -x^2) and C is given by r(t) = (t^3, t^2), 0 less than or equal to t less than or equal to 1. Evaluate the line integral integral_c (x+2y) dx+x^2 dy, where C consists of line segments from (0, 0) to (2, 1) and from (2, 1) to (3, 0). Evaluate the line integral integral_c xye^yz dy, where C is parameterized by x = t, y = t^2, z = t^3, 0 less than or equal to t less than or equal to 1. If C is a smooth curve given by a vector function r(t). a less than or equal to t less than or equal to b, show that integral_c r.dr = 1/2 [||r(b)||^2 - ||r(a)||^2]

Explanation / Answer

1)x=rcost y=rsint

x2+y2=r2

x2+y2=16

r2=16

=>r =4

x=4cost y=4sint

right half of circle ==> -/2<=t<=/2

r(t)=<4cost,4sint>, -/2<=t<=/2

F=xy4

F(r(t))=(4cost)(4sint)4=1024sin4t cost

r'(t)=<-4sint,4cost>

|r'(t)|=[(-4sint)2+(4cost)2]

|r'(t)|=4

Cxy4ds

=[-/2 to /2]F(r(t)) *|r '(t)| dt

=[-/2 to /2]1024sin4t cost *4 dt

=[-/2 to /2]4096sin4t cost dt

=[-/2 to /2](4096/5)sin5t

=(4096/5)sin5(/2) -(4096/5)sin5(-/2)

=(4096/5) -(-4096/5)

=8192/5

=1638.4

Cxy4ds =1638.4

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