You should graph the 3-petalled rose curve in the plane to establish the bounds of integration. Although the period of cos3Θ is 2π/3, we can see from the graph that if we let Θ range from 0 to π the whole graph is drawn. So we can set those as the bounds of integration.
For a polar area, we use the formula A =1/2 ∫ab r2dΘ:
A = 1/2 ∫0π (4cos(3θ))2dΘ = 8 ∫0π (cos2(3θ))dΘ . Now we need a double-angle trig substitution to make the function integrable: cos2(3θ) = 1/2(cos(6Θ) + 1)
A = 4 ∫0π (cos6Θ + 1)dΘ = 4(1/6sin6Θ + Θ)]0π = 4π
Tanvi ..
hello sir, i just have one question. why did you take range as 0 to pi? seeing the graph i thought it should've been 0 to 2pi, since the graph occupies all that area.10/24/22