Question

高数，计算曲面积分的一道题，谢谢啦~

Answer 1

令Dyz是曲面∑在yoz平面的投影，即{(y,z)|y^2+z^2<=4}
dx/dy=-2y，dx/dz=-2z
原式=∫∫(Dyz) (y^2+z^2)*√(1+4y^2+4z^2)dydz
=∫(0,2π)dθ∫(0,2)r^3*√(1+4r^2)dr
=2π*∫(0,2)r^3*√(1+4r^2)dr
令r=tant/2，则dr=sec^2t/2dt
原式=(π/16)*∫(0,arctan4) tan^3t*sec^3tdt
=(π/16)*∫(0,arctan4) (sec^2t-1)*sec^2t*d(sect)
=(π/16)*∫(0,arctan4) (sec^4t-sec^2t)d(sect)
=(π/16)*[(1/5)*sec^5t-(1/3)*sec^3t]|(0,arctan4)
=(π/16)*[(782/15)*√17+2/15]
=(π/120)*(391√17+1)