Question

Find the area of the region bounded by the curves 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

Check the picture below.

let's check where the quadratic and x = 6 meet, bear in mind we're integrating by "y", so

$\bf y=sin^{-1}\left( \frac{x}{6} \right)\implies sin(y)=\cfrac{x}{6}\implies \boxed{6sin(y)=x} \\\\\\ \begin{cases} 6sin(y)=x\\ x=6\\ y=0\\ x=0\leftarrow y-axis \end{cases}\\\\ -------------------------------\\\\ 6sin(y)=6\implies sin(y)=\cfrac{6}{6}\implies sin(y)=1\implies \measuredangle \boxed{y=\frac{\pi }{2}} \\\\\\ 6sin(y)=0\implies sin(y)=0\implies \to \boxed{\measuredangle y=0\ ,\ \pi }$

$\bf -------------------------------\\\\ \displaystyle \int\limits_{0}^{\frac{\pi }{2}}\ [6-6sin(y)]dy\implies \left. \cfrac{}{} 3y^2+6cos(y) \right]_{0}^{\frac{\pi }{2}} \\\\\\ \left[ \cfrac{3\pi 2}{4}+0 \right]-[0+6]\implies \cfrac{3\pi^2}{4}-6\approx 1.4022033$