Question

高等数学微分、二重积分的问题，求着两道题的详解，要步骤！

Answer 1

13
x^2+y^2<=2^(1/2)
[x^2+y^2]<=1
0<=x^2+y^2<1时,[x^2+y^2+1]=1
1<=x^2+y^2<=2^(1/2)时,[x^2+y^2+1]=2
定义D1={(x,y)|0<=x^2+y^2<1, 0<=x, 0<=y}
D2={(x,y)|1<=x^2+y^2<=2^(1/2), 0<=x, 0<=y}
积分=D1上的积分+D2 上的积分
=D1上(xy)的积分 + D2上(2xy)的积分
=D上(2xy)的积分-D1上(xy) 的积分
=S_{x:0->2^(1/2), y:0->[2-x^2]^(1/2)}(2xy)dxdy - S_{x:0->1, y:0->(1-x^2)^(1/2)}(xy)dxdy
=S_{x:0->2^(1/2)}x*(2-x^2)dx - S_{x:0->1}x*(1-x^2)/2dx
=[x^2-x^4/4]|_{x:0->2^(1/2)} - [x^2/4 - x^4/8]|_{x:0->1}
=[2-4/4] - [1/4 - 1/8]
=1 - 1/8
=7/8
14
(y-1)dx- xydy=0
(y-1)dx=xydy
dx/x = ydy/(y-1)=(y-1+1)dy/(y-1)=dy + dy/(y-1)
ln|x| = y + ln|y-1| + C
x=c(y-1)e^y

Answer 2

（1）我理解题意：[1+X^2+Y^2] 是个变量 诡异 求教
（2）dy/dx=(y-1)/xy=(1-1/y)/x
dy/(1-1/y)=dx/x
∫ (y/y-1)dy = ∫1/xdx
∫ [1+1/(y-1)]dy=∫1/xdx
y+lnly-1l=lnlxl +C