Last integration question in the book

[Hypochondriac]

its the last question in the last book so its lengthy, and not specific to one thing,

anyway a curve (that looks like an infinity symbol) is given by
x = 3cost and y = 9sin2t, 0≤t<2pi

a) find the cartesian in form y^2 = f(x)

i done that no problem and got y^2 = 4x^2(9-x^2)

b) show the shaded area enclosed by the curve and the x-axis is given by
\int_{0}^{\frac{\pi}{2}}Asin2t sint dt
stating the value of A

the infinity sign shaped curve is cut in half horizontally by the x axis, (and vertically by the y and by itself) the shaded sector is the top right one.

as the area = ∫ydx
y = 9sin2t

dx/dt = -3sint
so dx = -3sintdt

therefore area = ∫(9sin2t)(-3sintdt)
= ∫ -27sin2tsintdt

so A = -27
but the answer gives +27...

if i took +27 as my answer i can get the rest of the question right, but i get minus which doesn't make sense

if you need the diagram clarified i can draw it if you wish

thanks in advance

 

[Office_Shredder]

To find area, you should always take the absolute value of ydx (so if it goes under the curve, for example, it doesn't mess you up).

The reason you're gettting a negative sign here is because at t=0, x=3. At t=Pi/2, x=0. So basically, you're integrating in the negative direction. Going in the opposite direction would give you the positive answer you would expect (since it is all above the x-axis anyway)

 

[Hypochondriac]

ahhhh i get it so what I've found was
\int_{\frac{\pi}{2}}^{0}-27sin2t sint dt

which is equal to
\int_{0}^{\frac{\pi}{2}}-(-27)sin2t sint dt

which ofcourse is +27

thankyou so much that was really bugging me