# Simple intergral using parametric equations

## Homework Statement

i. $$x = 3cost,$$
ii. $$y = 9sin2t,$$
iii. $$0\leq t < 2\pi$$

iv.$$\int_0^\frac{\pi}{2} Asin2tsint \ dt$$

2. The attempt at a solution

So this is what I am given and I am supposed to be able to show that this is the integral for the shadded area between the curve and the x-axis in the 1st quadrant (sorry I know that relys on a graph, but the question is related I have isnt related to the actually shadded area anyway), so I can show the the interval is correct and form the intergral using

$$\int y \frac{dx}{dt} dt$$

but they state as answer that A is 27. Now I know that that must be the answer because the shaded region is above the x-axis, so the integral in those limits should be possitive, but whenever I try to from the intergral equation I get -27 becuase dx/dt is -3sint. Can anyone help me to see how I have gone wrong, thanks.

## Answers and Replies

Ah I have figured out the answer, basically the curve cuts the x-axis at 0 and 3 (well at least the shaded region), and getting the paramater t for each value is x=0, t= pi/2 and x = 3, t=0. so in actuall fact on initial formualtion of the intergral I should get:

$$\int_{\frac{\pi}{2}}^0 -27sin2tsint \ dt$$

and then if I reverse the limits of intergration of the intergral i get :

$$\int_0^{\frac{\pi}{2}} 27sin2tsint \ dt$$

Thanks all who did look (or will) :-)