Question

计算二重积分，D是由y=x，y=x^2围成，计算ffdxdy/√(x^2+y^2)

Answer 1

y=x与y=x^2的交点坐标是(0,0)及(1,1)
I=∫(0->1)dx∫(x^2->x)1/√(x^2+y^2)dy
=∫(0->1)dx∫(x^2->x)dy/x√[1+(y/x)^2]
=∫(0->1)dx/x^2∫(x^2->x)d(y/x)/√[1+(y/x)^2]
=∫(0->1)dx/x^2∫(x^2->x)darctan(y/x)
=∫(0->1)dx/x^2 arctan(y/x)|(x^2->x)
=∫(0->1)dx/x^2 (arctan1-arctanx)
=∫(0->1)dx/x^2(π/4-arctanx)
=π/4∫(0->1)dx/x^2-∫(0->1)dxarctanx/x^2
=π/4∫(0->1)d(-1/x) -∫(0->1)arctanxd(-1/x)
=-π/4∫(0->1)d(1/x) +∫(0->1)arctanxd(1/x)
=-π/4∫(0->1)d(1/x) +arctanx/x |(0->1) -∫(0->1)1/x*1/(x^2+1)dx
=-π/4∫(0->1)d(1/x) +arctanx/x |(0->1) -∫(0->1)xdx/x^2(x^2+1)
=-π/4∫(0->1)d(1/x) +arctanx/x |(0->1) -1/2∫(0->1)dx^2[1/x^2-1/(x^2+1)]
=-π/4∫(0->1)d(1/x) +arctanx/x |(0->1) -1/2∫(0->1)dx^2/x^2+1/2∫(0->1)d(x^2+1)/(x^2+1)
=-π/4∫(0->1)d(1/x) +arctanx/x |(0->1) -1/2lnx^2|(0->1) +1/2ln(x^2+1)|(0->1)
下面的解不出来了，最好用极坐标法了。nsjiang1 都解出来了，我就不再用了

Answer 2

用极坐标，θ:(0, π/4) r: (0,sinθ/cos²θ),代入：
=∫(0, π/4)dθ∫(0,sinθ/cos²θ)dr
=∫(0, π/4)(sinθ/cos²θ)dθ
={1/cosθ)|(0, π/4)
=√2-1