# Math

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This image has been Flagged as inappropriate Click to unflag You need to evaluate the area under the given curves, such that:

A = int_0^(pi) y*(dx)/(dt)

You need to evaluate (dx)/(dt) , such that:

(dx)/(dt) = (d(k(cos t + tsin t)))/(dt)

(dx)/(dt) = k(-sin t + sin t + tcos t)

(dx)/(dt) = k(t*cos t) => dx = k(t*cos t)*dt

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You need to evaluate the area under the given curves, such that:

A = int_0^(pi) y*(dx)/(dt)

You need to evaluate (dx)/(dt) , such that:

(dx)/(dt) = (d(k(cos t + tsin t)))/(dt)

(dx)/(dt) = k(-sin t + sin t + tcos t)

(dx)/(dt) = k(t*cos t) => dx = k(t*cos t)*dt

A = int_0^(pi) k(sin t - t cos t)k(t*cos t)*dt

Taking out the constant k^2 yields:

A = k^2*int_0^(pi) t*sin t*cos t dt - k^2*int_0^(pi) t^2 cos^2 t dt

You need to evaluate the integral k^2*int_0^(pi) t*sin t*cos t dt using the double angle identity, such that:

2sin t*cos t = sin (2t)

k^2/2*int_0^(pi) t*2*sin t*cos t dt = k^2/2*int_0^(pi) t*sin (2t) dt

You need to use integration by parts, such that:

int_0^(pi) t*sin (2t) dt = -t*(cos(2t))/2|_0^pi + (1/2)int_0^(pi)(cos(2t))

u = t => du = dt

dv = sin (2t) => v = -(cos(2t))/2

int_0^(pi) t*sin (2t) dt = -t*(cos(2t))/2|_0^pi + (sin (2t))/4|_0^pi

int_0^(pi) t*sin (2t) dt = -pi/2

You need to evaluate the integral k^2*int_0^(pi) t^2 cos^2 t dt using the half angle identity, such that:

cos^2 t = (1 + cos 2t)/2

k^2*int_0^(pi) t^2 cos^2 t dt = k^2/2*int_0^(pi) t^2(1 + cos 2t) dt

k^2*int_0^(pi) t^2 cos^2 t dt = k^2/2*int_0^(pi) t^2 dt + k^2/2*int_0^(pi) t^2*cos 2t dt

You need to use integration by parts for int_0^(pi) t^2*cos 2t dt , such that:

u = t^2 => du = 2tdt

dv = cos 2t => v = (sin 2t)/2

int_0^(pi) t^2*cos 2t dt = t^2/2*(sin 2t)|_0^pi - int_0^(pi) t(sin 2t)

A = k^2/2*int_0^(pi) t*sin (2t) dt + k^2/2*int_0^(pi) t^2 dt - k^2/2*int_0^(pi) t^2*cos 2t dt

A = k^2/2*int_0^(pi) t*sin (2t) dt + k^2/2*int_0^(pi) t^2 dt - k^2t^2/2*(sin 2t)|_0^pi + k^2/2*int_0^(pi) t*sin (2t) dt

A = -pi/2*k^2 + k^2/2*t^3/3|_0^pi

A =k^2/2(-pi + pi^3/3) => A = k^2(pi^3 - 3pi)/2

Hence, evaluating the area under the given parametric curves, yields A = k^2(pi^3 - 3pi)/2 .