Inverse Trig Question - Cambridge 3u (1 Viewer)

butterfly12345 · Nov 30, 2014

Consider the region bounded by y=[sin inverse x], the y-axis and the tangent to the curve at the point (sqrt(3)/2, pi/3)

Show that the area of the region is 1/4 units squared.
Show that the volume of the solid formed when the region is rotated about the y-axis is [pi/24(9sqrt3-4pi) unit^3

thanks

FrankXie · Dec 1, 2014

butterfly12345 said:
Consider the region bounded by y=[sin inverse x], the y-axis and the tangent to the curve at the point (sqrt(3)/2, pi/3)

Show that the area of the region is 1/4 units squared.
Show that the volume of the solid formed when the region is rotated about the y-axis is [pi/24(9sqrt3-4pi) unit^3

thanks

$$ Drawing a neat diagram, the desired area $ A= $ area of the triangle enclosed by the tangent, the y-axis and the line $ y=\frac{\pi}{3} $ minus the area of the region bounded by $ x=\sin y, $ the the y-axis and the line $ y=\frac{\pi}{3}$

$$ Therefore $ A=\frac12\times\frac{\sqrt3}{2}\times\sqrt3-\int_0^{\frac{\pi}{3}}\sin ydy=\frac34-\frac12=\frac14$

$$ Likewise, the desired volume equals the volume of the cone (which is formed by rotating the triangle) minus the volume of the solid formed by rotating $ x=\sin y$

$$ So $ V=\frac13\pi\left(\frac{\sqrt3}{2}\right)^2\sqrt3-\pi\int_0^{\frac{\pi}{3}}(\sin y)^2dy=\frac{\pi\sqrt3}{4}-\frac{\pi}{2}\int_0^{\frac{\pi}{3}}(1-\cos2y)dy=\frac{\pi\sqrt3}{4}-\frac{\pi}{2}\left(\frac{\pi}{3}-\frac{\sqrt3}{4}\right)=\frac{\pi\left(9\sqrt3-4\pi\right)}{24}$

Inverse Trig Question - Cambridge 3u (1 Viewer)

butterfly12345

New Member

FrankXie

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)