A. usingsymmetry, the area is just
∫[0,π/4] 2cosx - secx dx
= √2 - 2tanh-1 π/8
Using discs of thickness dx for the volume,
v = 2∫[0,π/4] π(R^2-r^2) dx
where R=2cosx and r=secx
v = 2∫[0,π/4] π((2cosx)^2-(secx)^2) dx = π^2
Using shells of thickness dy, we have to split the region where the curves intersect.
v = 2∫[1,√2] 2πrh dy
where r=y and h=x=arcsec(y)
+ 2∫[√2,2] 2πrh dy
where r=y and h=x=arccos(y/2)
v = 2∫[1,√2] 2πy*arcsec(y) dy
+ 2∫[√2,2] 2πy*arccos(y/2) dy
= π^2-2π + 2π
= π^2
A.Find the area of the region bounded above by y=2cosx and above by y=secx,-π/4≤x≤π/4. B.Find the volume of the sold generated by revolving the region in (A) above about the x-axis
2 answers
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