Consider S(1 to e)S(ln(x) to 1) sin(y^2)/x dydx?

Sketch the region of integration and then evaluate the integral.

b). Consider S(-2 to 2)S(0 to sqrt(4-y^2))S((x^2+y^2)/2 to 2) sqrt(x^2+y^2) dz dx dy

by firstly converting the integral to an equivalent integral in cylindrical coordinates.

Comments

  • a) The region is bounded by y = ln x to y = 1 with x in [1, e].

    Rewrite this as x = 0 to x = e^y with y in [0, 1].

    So, the integral equals

    ∫(y = 0 to 1) ∫(x = 1 to e^y) sin(y^2) * 1/x dx dy

    = ∫(y = 0 to 1) sin(y^2) * ln |x| {for x = 1 to e^y} dy

    = ∫(y = 0 to 1) sin(y^2) * y dy

    = (-1/2) cos(y^2) {for y = 0 to 1}

    = (1/2) (1 - cos 1).

    -----------------

    b) Note that the xy-region is the right half of the circle x^2 + y^2 = 4.

    So, we obtain via cylindrical coordinates:

    ∫(θ = -π/2 to π/2) ∫(r = 0 to 2) ∫(z = r^2/2 to 2) r * (r dz dr dθ)

    = π ∫(r = 0 to 2) r^2 (2 - r^2/2) dr

    = π ∫(r = 0 to 2) (1/4)(4r^2 - r^4) dr

    = (π/4)(4r^3/3 - r^5/5) {for r = 0 to 2}

    = 16π/15.

    I hope this helps!

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