向量场与多元微积分习题解析-优快云博客

本文链接：https://blog.youkuaiyun.com/Yjl_Richard/article/details/116567139

本文详细解答了多个向量场的散度、旋度计算问题，涉及多元函数的微积分应用，包括保守场判断和有势场的势函数求解。同时还探讨了相关数学概念如梯度、旋度和散度的性质。

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                    求下列向量场在指定点的散度：
 (1)  F = ( 5 x y , x y z , 3 l n ( x z ) ) \bm{F}=(5xy,\frac{xy}{z},3ln(xz)) F=(5xy,zxy​,3ln(xz))，在 ( 6 , 1 , 3 ) (6,1,3) (6,1,3)处
 (2)  F = x 2 + y 2 i + l n ( y + y 2 + z 2 ) j + y 2 + z 2 k \bm{F}=\sqrt{x^2+y^2}\bm{i}+ln(y+\sqrt{y^2+z^2})\bm{j}+\sqrt{y^2+z^2}\bm{k} F=x2+y2 
             
            ​i+ln(y+y2+z2 
             
            ​)j+y2+z2 
             
            ​k，在 ( 3 , 4 , 3 ) (3,4,3) (3,4,3)处
 解：
 (1)  d i v F = ∂ ( 5 x y ) ∂ x + ∂ ( x y z ) ∂ y + ∂ ( 3 l n ( x z ) ) ∂ z = 5 y + x z + 3 z div\bm{F}=\frac{\partial(5xy)}{\partial x}+\frac{\partial(\frac{xy}{z})}{\partial y}+\frac{\partial(3ln(xz))}{\partial z}=5y+\frac{x}{z}+\frac{3}{z} divF=∂x∂(5xy)​+∂y∂(zxy​)​+∂z∂(3ln(xz))​=5y+zx​+z3​
  d i v F ( 6 , 1 , 3 ) = 8 div\bm{F}_{(6,1,3)}=8 divF(6,1,3)​=8
 (2)  d i v F = ∂ ( x 2 + y 2 ) ∂ x + ∂ ( l n ( y + y 2 + z 2 ) ) ∂ y + ∂ ( y 2 + z 2 ) ∂ z = x x 2 + y 2 + 1 + z y 2 + z 2 div\bm{F}=\frac{\partial(\sqrt{x^2+y^2})}{\partial x}+\frac{\partial(ln(y+\sqrt{y^2+z^2}))}{\partial y}+\frac{\partial(\sqrt{y^2+z^2})}{\partial z}=\frac{x}{\sqrt{x^2+y^2}}+\frac{1+z}{\sqrt{y^2+z^2}} divF=∂x∂(x2+y2 
                     
                    ​)​+∂y∂(ln(y+y2+z2 
                     
                    ​))​+∂z∂(y2+z2 
                     
                    ​)​=x2+y2 
                     
                    ​x​+y2+z2 
                     
                    ​1+z​
  d i v F ( 3 , 4 , 3 ) = 7 5 div\bm{F}_{(3,4,3)}=\frac{7}{5} divF(3,4,3)​=57​
设 a \bm{a} a为常向量， r = ( x , y , z ) , r = ∣ r ∣ \bm{r}=(x,y,z),r=|\bm{r}| r=(x,y,z),r=∣r∣， f ( r ) f(r) f(r)为r的可微函数，求下列各式的值：
 (1)  ∇ ⋅ ( r a ) ; \nabla\cdot(r\bm{a}); ∇⋅(ra);(2)  ∇ ⋅ ( r 2 a ) ; \nabla\cdot(r^2\bm{a}); ∇⋅(r2a);(3)  ∇ ⋅ ( r n a ) \nabla\cdot(r^n\bm{a}) ∇⋅(rna)(n为正整数);(4)  ∇ ⋅ ( r r 3 ) ; \nabla\cdot(\frac{\bm{r}}{r^3}); ∇⋅(r3r​);
 (5)  ∇ ⋅ [ f ( r ) a ] ; \nabla\cdot[f(r)\bm{a}]; ∇⋅[f(r)a];(6)  ∇ ⋅ [ g r a d f ( r ) ] ; \nabla\cdot [gradf(r)]; ∇⋅[gradf(r)];(7)  ∇ ⋅ [ f ( r ) r ] \nabla\cdot[f(r)\bm{r}] ∇⋅[f(r)r]
 解：
 设 a = ( C 0 , C 1 , C 2 ) \bm{a}=(C_0,C_1,C_2) a=(C0​,C1​,C2​)， ∣ r ∣ = x 2 + y 2 + z 2 |r|=\sqrt{x^2+y^2+z^2} ∣r∣=x2+y2+z2 
             
            ​
 (1)  ∇ ⋅ ( r a ) = ∂ ( C 0 x 2 + y 2 + z 2 ) ∂ x + ∂ ( C 1 x 2 + y 2 + z 2 ) ∂ y + ∂ ( C 2 x 2 + y 2 + z 2 ) ∂ z = C 0 x + C 1 y + C 2 z x 2 + y 2 + z 2 = a ⋅ r r \nabla\cdot(r\bm{a})=\frac{\partial(C_0\sqrt{x^2+y^2+z^2})}{\partial x}+\frac{\partial(C_1\sqrt{x^2+y^2+z^2})}{\partial y}+\frac{\partial(C_2\sqrt{x^2+y^2+z^2})}{\partial z}=\frac{C_0x+C_1y+C_2z}{\sqrt{x^2+y^2+z^2}}=\frac{\bm{a}\cdot\bm{r}}{r} ∇⋅(ra)=∂x∂(C0​x2+y2+z2 
                     
                    ​)​+∂y∂(C1​x2+y2+z2 
                     
                    ​)​+∂z∂(C2​x2+y2+z2 
                     
                    ​)​=x2+y2+z2 
                     
                    ​C0​x+C1​y+C2​z​=ra⋅r​
 (2)  ∇ ⋅ ( r 2 a ) = ∂ ( C 0 ( x 2 + y 2 + z 2 ) ) ∂ x + ∂ ( C 1 ( x 2 + y 2 + z 2 ) ) ∂ y + ∂ ( C 2 ( x 2 + y 2 + z 2 ) ) ∂ z = 2 C 0 x + 2 C 1 y + 2 C 2 z = 2 a ⋅ r \nabla\cdot(r^2\bm{a})=\frac{\partial(C_0(x^2+y^2+z^2))}{\partial x}+\frac{\partial(C_1(x^2+y^2+z^2))}{\partial y}+\frac{\partial(C_2(x^2+y^2+z^2))}{\partial z}=2C_0x+2C_1y+2C_2z=2\bm{a}\cdot\bm{r} ∇⋅(r2a)=∂x∂(C0​(x2+y2+z2))​+∂y∂(C1​(x2+y2+z2))​+∂z∂(C2​(x2+y2+z2))​=2C0​

高数习题8.7