Question

Find f · dr c for the given f and

c. f = x2 i + y2 j and c is the top half of a circle of radius 3 starting at the point (3, 0) traversed counterclockwise.

Answer 1

The question involves computing the line integral of a vector field along a curve, specifically the upper semicircle of radius 3.

Step-by-step explanation:

To find the line integral of f · dr along the curve C, we first need to parameterize the curve C. Since C is the top half of a circle of radius 3 starting at the point (3, 0) and traversed counterclockwise, we can parameterize it as follows:

x = 3 + 3 cos(t)

y = 3 sin(t)

where 0 ≤ t ≤ π.

Next, we can calculate the dot product of f and dr:

f · dr = (x² i + y² j) · (-3 sin(t) i + 3 cos(t) j) = -3x sin(t) + 3y cos(t)

Substituting the parameterization of C into this expression, we get:

f · dr = -9sin(t)cos(t) + 9sin²(t)

Finally, we can evaluate the line integral using the following formula:

∫C f · dr = ∫aᵇ f(r(t)) · r'(t) dt

where r(t) is the parameterization of the curve C and a and b are the start and end points of the curve. In this case, we have:

∫C f · dr = ∫0^π (-9sin(t)cos(t) + 9sin²(t)) dt

Evaluating the integral, we get:

∫C f · dr = 9/2π

Answer 2

Parameterize the path
`\mathcal C` by


`\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j`

with


`x(t)=3\cos t`

`y(t)=3\sin t`

and
`0\le t\le\pi`. Then


`\displaystyle\int_(\mathcal C)\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_(t=0)^(t=\pi)\mathbf f(x(t),y(t))\cdot(\mathrm d\mathbf r)/(\mathrm dt)\,\mathrm dt`

`=\displaystyle\int_0^\pi(9\cos^2t\,\mathbf i+9\sin^2t\,\mathbf j)\cdot(-3\sin t\,\mathbf i+3\cos t\,\mathbf j)\,\mathrm dt`

`=27\displaystyle\int_0^\pi(\cos t\sin^2t-\sin t\cos^2t)\,\mathrm dt=-18`