Hi,
   I just finished a math quiz. I needed to find the length of the curve and area of the function, r=3*cos(theta)-2*sin(theta) bounded between 0<=Pi<=2*Pi.
   On the quiz I used Area=int(1/2*(r^2)) dtheta. For the length of the curve I used L=int(sqrt(r^2 + r'(theta)) dtheta.

How do you plug this into Maple and get an answer?
I came up with 20.42.... sq units for the area and 22.65.... for the length of the curve.

Thank you,
Jay.

Hi,

1st post. I'm trying to integrate the following function:

h:=t->(2*t-1)*cos*sqrt(3*(2*t-1)^2+6)/(sqrt(3*(2*t-1)^2+6));

h:=t->(2*t-1)*cos*(sqrt(3*(2*t-1)^2+6))/(sqrt(3*(2*t-1)^2+6));

h:=t->(2*t-1)*(cos*(sqrt(3*(2*t-1)^2+6)))/(sqrt(3*(2*t-1)^2+6));

h:=t->((2*t-1)*(cos*(sqrt(3*(2*t-1)^2+6))))/(sqrt(3*(2*t-1)^2+6));

int(h(t),t); ** Integration command. I've also replaced the "h(t)" with the entire function.

I've tried the following:

1. Changed the "t" to "x" throughout function.

2. Added parens around sqrt portion.

3. Added parens to include "cos" and then added to include the beginning (2*t-1).

4. I've added brackets around the numerator but this just causes Maple to reprint the function with the inegration sign in front of the function.

5. I've also tried using the Integration tutor. It returns that maple is unable to calculate.

6. Repeat all the above in Maple 2015, same answer.

I always get cos(t^2-t).

The math book claims the answer is 1/6*sin*sqrt(3*(2*t-1)^2+6). When I perform the inegration on paper I get the same answer.

Any suggestions or corrections would be great.

Thank you,

Jay.