Answer:
a) [tex]5xz + xy + 4yz[/tex]
b) 10
Step-by-step explanation:
a) Here [tex]F(x,y,z)=(5z+y)i+(4z+x)j+(4y+5x)k[/tex]
Since the case [tex]F[/tex] = ∇[tex]f[/tex] holds, then
∇[tex]f = f_xi+f_yj+f_zk[/tex] = [tex](5z+y)i+(4z+x)j+(4y+5x)k[/tex]
So, [tex]f_x = 5z + y[/tex]
If we integrate [tex]f_x[/tex] with respect to x, we will get an integration constant C which is also a function that depends to y and z.
Hence,
[tex]f = \int f_xdx = 5xz + xy + g(y,z)[/tex]
Now we need to find g(y,z).
So first let's take the derivative of g(y,z) with respect to y.
[tex]f_y = x + g_y(y,z) = 4z + x[/tex]
Hence, [tex]g_y(y,z) = 4z[/tex]
So now, if we integrate [tex]g_y[/tex] with respect to y to find g(y,z)
[tex]g = \int g_ydy = 4yz + C[/tex]
Thus,
[tex]f = 5xz + xy + g(y,z) = 5xz + xy + 4yz + C[/tex]
And since [tex]f(0,0,0)=0[/tex], then [tex]C=0[/tex]
Thus,
[tex]f = f(x,y,z) = 5xz + xy + 4yz[/tex]
b) By the Fundamental Theorem of Line Integrals, we know that
[tex]\int\limits^a_b F. dr = F[r(b)]-F[r(a)][/tex]
Hence,
[tex]\int\limits^a_b F. dr = F(1,1,1)-F(0,0,0) =[(5+1+4)-(0+0+0)]=10[/tex]
