Question

Find the area of the region bounded by the courses y = sin^-1(x/6), y = 0, and x = 6 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative.

Answer 1

Given

The equation is given

`y=sin^(-1)((x)/(6))`

y=0 and x=6.

Required

To determine the area of region bounded by courses of the equation given.

Step-by-step explanation
`\begin{gathered} y=sin^(-1)((x)/(6))\Rightarrow\Rightarrow siny=(x)/(6) \\ 6siny=x \end{gathered}`
`x=6,y=0`

Therefore

`\begin{gathered} 6siny=6 \\ siny=1 \\ y=(\pi)/(2) \end{gathered}`
`\begin{gathered} 6siny=0 \\ siny=0 \\ y=0,\pi \end{gathered}`

Now integrate the equation.

`\int_0^{(\pi)/(2)}(6-6siny)dy=(6y+6cosy)_0^{(\pi)/(2)}`
`((6\pi)/(2)+6cos(\pi)/(2))-(6*0+6cos0)=3\pi-6`Answer

Hence the area of region bounded by the courses is

`3\pi-6`