Answer to Question #275333 in Calculus for Good heart

Question

I. Change the following point from polar to rectangular coordinate.

(3/2, n/12)

II. Evaluate f²(3ײ- 6x ‐2)dx

Answer 1

a) $x=r\cos \theta, y=r\sin \theta$

$$x=\dfrac{3}{2}\cos\dfrac{\pi}{12}=\dfrac{3}{2}(\dfrac{\sqrt{6}+\sqrt{2}}{4})=\dfrac{3(\sqrt{6}+\sqrt{2})}{8}$$ $$y=\dfrac{3}{2}\sin\dfrac{\pi}{12}=\dfrac{3}{2}(\dfrac{\sqrt{6}-\sqrt{2}}{4})=\dfrac{3(\sqrt{6}-\sqrt{2})}{8}$$ $$\big(\dfrac{3(\sqrt{6}+\sqrt{2})}{8}, \dfrac{3(\sqrt{6}-\sqrt{2})}{8}\big)$$

b)

$$\int (3x^2-6x-2)dx=x^3-3x^2-2x+C$$