Computing area with parametrically represented boundaries : If the boundary of a figure is represented by parametric equation, i.e., `x=x(t), y=(t),` then the area of the figure is evaluated by one of the three formulas :
`S=-int_(alpha)^(beta)y(t)x'(t)dt,`
`S=int_(alpha)^(beta)x(t)y'(t)dt,`
`S=(1)/(2)int_(alpha)^(beta)(xy'-yx')dt,`
Where `alpha and beta` are the values of the parameter t corresponding respectively to the beginning and the end of the traversal of the curve corresponding to increasing t.
The area of the loop described as
`x=(t)/(3)(6-t),y=(t^(2))/(8)(6-t)` is

Options

A. `(27)/(5)` sq. units
B. `(24)/(5)` sq. units
C. `(27)/(6)` sq. units
D. `(21)/(5)` sq. units

Answer

A

Explanation

`overset(6)underset(0)int((3)/(2)t^(2)-(1)/(2)t^(3)+(1)/(24)t^(4))dt=(3)/(2)xx(6^(3))/(3)xx(1)/(2)(6^(4))/(4)xx(1)/(24)(6^(5))/(5)`
`=(6^(3))/(2)-(6^(4))/(8)+(6^(4))/(20)`
`=6^(4)((1)/(12)-(1)/(8)+(1)/(20))`
`=(54)/(5)`.
`therefore" "(1)/(2)overset(6)underset(0)int(xy'-yx')dx=(1)/(2)xx(54)/(5)=(27)/(5)` sq. units.