Question

高数题最后的-√2怎么得到的

Answer 1

由对称性，得 I = 4 ∫4, π/2>dt ∫ √[1-(rsint)^2] rdr = 2 ∫4, π/2>dt ∫ [-(csct)^2]√[1-(rsint)^2] d [1-(rsint)^2] = (4/3) ∫4, π/2>dt [-(csct)^2][1-(rsint)^2]^(3/2) = (4/3) ∫4, π/2> (csct)^2 [1-(cost)^3)] dt = (4/3) ∫4, π/2> [1-(cost)^3)]dt/[1-(cost)^2] = (4/3) ∫4, π/2> [1+cost+(cost)^2)]dt/(1+cost) = (4/3) ∫4, π/2> [cost+1/(1+cost)] dt = (4/3)[sint + csct - cott]4, π/2> = 4-2√2