作业帮几道高数题,马上要补考了,
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/16 18:05:22
几道高数题,
马上要补考了,
马上要补考了,
三、dz/dx=fx+fydy/dx
e^(x/y)[(1/y)+(-x/y^2)]dy/dx=(dy/dx)/z+(-y/z^2)dz/dx
故：fydy/dx -dz/dx=-fx
[x(z^2)e^(x/y)+(y^2)z]dy/dx-(y^3)dz/dx=y(z^2)e^(x/y)
∴dz/dx={y(z^2)e^(x/y)fy+ [x(z^2)e^(x/y)+(y^2)z]fx}/{ [x(z^2)e^(x/y)+(y^2)z]-(y^3)fy}
----------------------------
四、S=2∫∫dS=2∫∫√[1+(∂z/∂x^2+(∂z/∂y)^2]dxdy=2∫∫(1/z)dxdy
=4∫(0,π/2)d θ∫(0,cos θ)[r/√(1-r^2)]dr=4∫(0,π/2)(1-sinθ)d θ=2π-4
e^(x/y)[(1/y)+(-x/y^2)]dy/dx=(dy/dx)/z+(-y/z^2)dz/dx
故：fydy/dx -dz/dx=-fx
[x(z^2)e^(x/y)+(y^2)z]dy/dx-(y^3)dz/dx=y(z^2)e^(x/y)
∴dz/dx={y(z^2)e^(x/y)fy+ [x(z^2)e^(x/y)+(y^2)z]fx}/{ [x(z^2)e^(x/y)+(y^2)z]-(y^3)fy}
----------------------------
四、S=2∫∫dS=2∫∫√[1+(∂z/∂x^2+(∂z/∂y)^2]dxdy=2∫∫(1/z)dxdy
=4∫(0,π/2)d θ∫(0,cos θ)[r/√(1-r^2)]dr=4∫(0,π/2)(1-sinθ)d θ=2π-4