# plz help to get the answer of the attcehd pic.

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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` .**

**Sources:**