calculus
posted by Naseba .
Let R be the region in the first quadrant that is enclosed by the graph of y = tanx, the xaxis, and the line x = π/3
h. Find the area of R
i. Find the volume of the solid formed by revolving R about the xaxis

what's the problem?
A = ∫[0,pi/3] tanx dx
integrating tanx is easy, since
tanx = sinx/cosx = d(cosx)/cosx
∫tanx dx = ln(cosx)
A = ln(cosx) [0,pi/3]
= ln(cos pi/3) + ln(cos 0)
= ln(1/2) + ln(1)
= ln2
v = ∫[0,pi/3] pi*r^2 dx
= pi*∫[0,pi/3] tan^2x dx
= pi*∫[0,pi/3](sec^2x  1) dx
= pi*(tanx  x) [0,pi/3]
. . .