Inverse Trigonometry Question. (1 Viewer)

[locked.on]

Stuck here.

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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)]

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Help appreciated, thanks in advanced.

 

[azureus88]

first part,

a)find derivative of inverse sine to find gradient at the point
b)form equation of tangent
c)find y intercept (when x=0)
d)find area of triangle with base sqrt(3) and height sqrt(3)/2
e)find area of part bounded by y=pi/3 and y=0 and inverse sine. (you could integrate sin[x] from x=0 to x=pi/3.
g)shaded area = area of triangle - area in part e)

second part is similar, except its volumes.