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Let R be the region in the first quadrant that is enclosed by the graph of y = tanx, the x-axis, 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 x-axis

  • calculus -

    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]
    . . .

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