Question

What is the area enclosed by `r=-9cos((5theta)/3-(pi)/8)+3sin((theta)/2+(3pi)/4)` between `theta in [0,pi/2]`?

Answer 1

Area `=19.9632" "`square units

Explanation:

The formula to obtain the area is
Area `=int_a^b 1/2r^2 d theta`

Our given equation
`r=-9 cos((5theta)/3-pi/8)+3*sin(theta/2+(3pi)/4)`

We square r first then transform the terms into some expressions that are integrable

`r^2=[-9 cos((5theta)/3-pi/8)+3*sin(theta/2+(3pi)/4)]^2`

`r^2=81 cos^2((5theta)/3-pi/8)-54 sin(theta/2+(3pi)/4)* cos((5theta)/3-pi/8)+9 sin^2(theta/2+(3pi)/4)`

Using Half-Angle Formulas and Product to Sum Formulas, both from Trigonometry, we transform into as follows

`r^2=45+81/2*cos((10 *theta)/3-pi/4)-27 sin((13*theta)/6+(5pi)/8)-27 sin((-7*theta)/6+(7pi)/8)-9/2*cos(theta+(3pi)/2)`

This is now integrable

Area `=1/2int_0^(pi/2)[45+81/2*cos((10 *theta)/3-pi/4)-27 sin((13*theta)/6+(5pi)/8)-27 sin((-7*theta)/6+(7pi)/8)-9/2*cos(theta+(3pi)/2)]d theta`

Area `=19.9632" "`square units

God bless....I hope the explanation is useful.