Verified answer

Green Theo ,

=INT_R ( -3x^2 -3y^2 ) dA

R is the region of a circle r=3

=-3 INT (x^2+y^2) dA

polar coord ( Counterclockwise) 0<T<2pi ) dA=+dA

=-3 INT 9 dA

=-27 INT (+dA)

=-27piR^2

=-243pi

we can rewrite the line imperative as: ?_C F•dr = (a to b) ? [F(r(t))•r'(t)] dt the following, r(t) is the line from (-a million,4) to (a million,2), and this may be expressed as: r(t) = <t , -t + 3>, the position t stages from -a million to at least a million (those might want to be the barriers of integration) and r'(t) = <-a million , -a million> for F(r(t)), all this suggests is that we replace the x and y parts in F with what x and y are given to be in r(t). all of us know that r(t) = <-t , -t + 3>. we can rewrite this as x = -t, y = -t + 3, and so all we do is change those into F, so F(r(t)) is: F(r(t)) = <2(-t)(-t + 3) , (-t)^2 - (-t + 3)> = <2t^2 - 6t , t^2 - t - 3> now, we take the dot made from F(r(t)) and r'(t): F(r(t))•r'(t) = <2t^2 - 6t , t^2 - t - 3>•<-a million , -a million> = -2t^2 + 6t - t^2 + t + 3 = -3t^2 + 7t + 3 now, only combine that over -a million ? t ? a million: (-a million to at least a million) ? [-3t^2 + 7t + 3] dt = [-t^3 + (7/2)t^2 + 3t] | (-a million to at least a million) = [-(a million)^3 + (7/2)(a million)^2 + 3(a million)] - [-(-a million)^3 + (7/2)(-a million)^2 + 3(-a million)] = [-(a million) + (7/2)(a million) + 3] - [-(-a million) + (7/2)(a million) - 3] = [-a million + (7/2) + 3] - [a million + (7/2) - 3] = [2 + (7/2)] - [-2 + (7/2)] = 2 + (7/2) + 2 - (7/2) = 4

First you need to convert the curve to parametric form:

x^2+y^2=9 => x=3cost, y=3sint, dx=3sint, dy=-3cost

Now you have the ∫y^3dy = ∫(3sint)^3*3sint dt

This integral evaluates to 3^4*1/32*[12x - 8sin2x + sin4x]

And you have -∫x^3dx = ∫(3cost)^3*-3cost dt

This evaluates to -3^4*1/32*[12x + 8sin2x + sin4x]

So your result is -3^4*1/32*[16sin2x + 2sin4x]

Hope that helps!