求微分方程xydx=√(1+x^2)dy的通解

z=cos(xy+y^2)()的全微分为()(2.0分)A.()dz=-sin(xy+y^2)(dx+dy)()B.()dz=-sin(xy+y^2)[ydx+(x+2y)dy]()C.()dz=-sin(xy+y^2)(y^2dx+xdy)()D.()dz=-sin(xy+y^2)[xydx+(x+y^2)dy]

计算\(\int_L {xydx} \)，其中\(L\) 是抛物线\(y^2=x\) 上从点\((1, - 1)\) 到点\((1,1)\) 的一段弧。 A: \({3 \over 4}\) B: \({1 \over 2}\) C: \({2 \over 3}\) D: \({4 \over 5}\)

计算 \(\oint_L {xydx} \)，其中\(L\) 为圆周 \({(x - a)^2} + {y^2} = {a^2}(a > 0)\)及 \(x\)轴所围成的在第一象限内的区域整个边界（按逆时针方向）. A: \({\pi \over 2}{a^3}\) B: \( - {\pi \over 3}{a^3}\) C: \( {\pi \over 3}{a^3}\) D: \( - {\pi \over 2}{a^3}\)

函数\(z = {\left( {xy} \right)^x}\)的全微分为 A: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + x{\left( {xy} \right)^x}dy\) B: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) C: \(dz = {\left( {xy} \right)^x}\ln xydx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) D: \(dz = {\left( {xy} \right)^x}\left( {1 + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\)