Calc 3 (parametric equations) My answers won't match/can't find arithmetic error

[d3nat]

two questions.
I know I'm doing the work right, but I can't get my answers to match and they are pretty close. I think it's just some arithmetic errors.
Help? I've been trying to solve my mistakes forever. I can't find them!

Homework Statement

Use the parametric equation of an ellipse x = acos(t) and y=bsin(t) from 0<t<2PI (those are or equal to signs)

Homework Equations

A=INTEGRAL from 0 - 2PI ydx

The Attempt at a Solution

INTEGRAL from 0-2PI bsin(t)(-asin(t) dt
= -ab INTEGRAL 0 -2PI sin^2(t) dt
= -ab EVALUATED 1/2 (t) - 1/4*sin(2t) from 0-2PI
= -ab ((1/2*(2PI) - 1/4 (sin(2*2PI) - 0)
= -abPI

But the answer is abPI. I don't know where the negative sign goes... or if I am supposed to have another one that cancels mine out

Homework Statement

Find the exact length of the curve

Homework Equations

x = 1+3t^2
y= 4+2t^3
0<t<1 (or equal)
)^2)
L= INTEGRAL from a - b SQUARE ROOT (dx/dt)^2 +(dy/dt)^2 END SQUARE ROOT dt

The Attempt at a Solution

dx/dt = 6t
dy/dt = 6t^2

L= INTEGRAL from 0-1 SQARE ROOT (6t)^2 + (6t^2)^2 END SQUARE ROOT dt
= 6 INTEGRAL from 0-1 t* SR 1+t^2 dt
using u substitution

u = 1+t^2
du = 2t dt
boundaries change to 1 - 2
= 3 * integral from 1-2 square root u du
= 3 * 2*u^3/2 evaluated from 1-2
= 3 * ( 2(2)^3/2 - 2(1)^3/2 )
= 3 * (4 square root (2) - 2)
= 12 square root (2) - 6


answer is supposed to be
4 square root (2) - 1

 

[d3nat]

Solved #2.
I was doing the integral wrong. DUH! I can't believe it took me so long to figure out. I was bringing 2 (reversed of 1/2) down, not 2/3 (reversed of 3/2) and I wrote the answer down wrong. It was supposed to be 4 square root (2) - 2.

So I solved that one after an 'aha!' moment. Still working on the first.

 

[flyingpig]

For the integral

$$-ab\int_{0}^{2Pi} \sin^2 t dt$$

use $$\sin^2 t = \frac{1 - \cos(2t)}{2}$$