says said:Homework Statement
Evaluate ∫c (x + y) ds, where C is the circle centred at (1/2, 0) with radius 1/2.
Homework Equations
The Attempt at a Solution
parametrise
x=1/2cos(t)
y=1/2sin(t)
0≤t≤2π
ds=√dx2+dy2
=√(1/2)2-sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2(sin2(t)+cos2(t))
ds=1/2
∫ [(1/2)cos(t) + (1/2)sin(t)]*(1/2) dt, where 0≤t≤2π
I evaluated this integral and got 0. Is this because C is a simple closed curve?
says said:The problem statement didn't mention the direction. So I just assume anti-clockwise.
Do you know where I have made my mistake? I have been on this problem for a while now and can't seem to find it. I've put the integral into a computation as well and got back 0 as the answer.
Since C is a circle, perhaps a more natural expression for ds would be r ⋅ dθsays said:Argh!
The circle is centred at (1/2,0) not at the origin. I'm not entirely sure how this changes anything though.
says said:I found a source online that suggested if the circle is not at the origin then:
x = h + r cos(t)
y = k + r sin(t)
so
x = 1/2 + (1/2 cos(t))
y = 1/2 sin(t)
edit: FOUND THE INFO I WAS LOOKING FOR!
Circle with Center at Point (h,k)
http://www.regentsprep.org/regents/math/geometry/gcg6/LCirh3.gif
where the center is (h,k)
and the radius is r
so
x= 1/2cos(t)-1/2
y= 1/2sin(t)
A line integral over a circle region is a mathematical concept that involves calculating the integral of a function along a curve that forms a closed loop around a circular region. The curve can be any smooth path, such as a straight line or a curved line, as long as it encloses the circular region.
The formula for calculating a line integral over a circle region depends on the type of line integral being used. For example, if a line integral is being calculated using the standard definition, the integral can be evaluated by dividing the circle into small segments and using the values of the function at each point along the curve. Alternatively, if a line integral is being calculated using the Green's theorem, the integral can be evaluated by converting it into a double integral over the enclosed region.
Line integrals over circle regions have various applications in physics, engineering, and mathematics. They are commonly used to calculate the work done by a force along a circular path, the circulation of a vector field around a circular region, and the flux of a vector field through a circular surface. They are also used in the study of fluid mechanics, electromagnetism, and complex analysis.
A conservative vector field is one in which the line integral along any closed curve is equal to zero. If a line integral over a circle region is calculated using a conservative vector field, the result will always be zero. This is because the starting and ending points of the curve are the same, and the integral cancels out. Therefore, line integrals over circle regions can be used to determine if a vector field is conservative or not.
Yes, there are some special properties of line integrals over circle regions. One of the most important is the fact that the value of the integral is independent of the path taken between the starting and ending points of the curve. This property is known as path independence and is a consequence of the fundamental theorem of calculus for line integrals. Additionally, line integrals over circle regions are also invariant under rotation, meaning that the value of the integral will remain the same if the circle is rotated around its center.