请教几个高数问题

再次请教几个高数题目！见附件，谢

解: 1。 原式=∫[∫y/(1+x^2+y^2)^(3/2)dy]dx =-∫[(1+x^2+y^2)^(-1/2)|]dx =-∫[(2+x^2)^(-1/2)-(1+x^2)^(-1/2)]dx =-[ln(x+√(2+x^2)-ln(x+√(1+x^2)] =ln[(√6+2√3-√2-2)/2] 2。 化为极坐标,设x=rcosθ,y=rsinθ,0≤r≤R,0≤θ≤2π 则原式=∫∫[(rcosθ)^2/a^2+(rsinθ)^2/b^2]rdrdθ =∫[∫[(cosθ)^2/a^2+(sinθ)^2/b^2]r^3dr]dθ =∫[[(cosθ)^2/a^2+(sinθ...全部

  解: 1。 原式=∫[∫y/(1+x^2+y^2)^(3/2)dy]dx =-∫[(1+x^2+y^2)^(-1/2)|]dx =-∫[(2+x^2)^(-1/2)-(1+x^2)^(-1/2)]dx =-[ln(x+√(2+x^2)-ln(x+√(1+x^2)] =ln[(√6+2√3-√2-2)/2] 2。
   化为极坐标,设x=rcosθ,y=rsinθ,0≤r≤R,0≤θ≤2π 则原式=∫∫[(rcosθ)^2/a^2+(rsinθ)^2/b^2]rdrdθ =∫[∫[(cosθ)^2/a^2+(sinθ)^2/b^2]r^3dr]dθ =∫[[(cosθ)^2/a^2+(sinθ)^2/b^2]r^4/4|]dθ =(R^4/4)∫[(cosθ)^2/a^2+(sinθ)^2/b^2]dθ =(R^4/4)[(θ/2+(1/4)sin2θ)/a^2+(θ/2-(1/4)sin2θ)/b^2]| =πR^2(a^2+b^2)/(4a^2b^2) 3。
   原式=∫[∫e^(-y^2)dx]dy =∫[xe^(-y^2)|]dy =∫[ye^(-y^2)]dy =(-1/2)e^(-y^2)| =(e-1)/(2e) 4。
   原式=∫[∫(x+y)dx]dy+∫[∫(x+y)dx]dy =∫[(x^2/2+xy)|]dy+∫[(x^2/2+xy)|]dy =∫[y^2]dy =(1/3)y^3| =1/3 。收起