$\int_C ( 2 xz \, \vec {\mathbf i} + x^2z \, \vec {\mathbf j}+ x^2y \, \vec {\mathbf k}) \mathrm d \, \vec {\mathbf r}$ : Line Integral

Question

Evaluate $\int_C \vec {\mathbf F} \mathrm d \, \vec {\mathbf r}$ , where $\vec {\mathbf F} (x,y,z) = 2 xz \, \vec {\mathbf i} + x^2z \, \vec {\mathbf j}+ x^2y \, \vec {\mathbf k}$ , and C is the path from (0 , 1 , 2) to (1 , 2, 3) that consists of three line segments parallel to the x -axis, y-axis, and z-axis, in this order.

Ill try it in a few moments

score 3 · Accepted Answer · answered Mar 09 '13 at 01:36

3

Put $\,C_i\,$ = the path for the variable $\,i\,$ . For example, the first one takes $\,y=1\,,\,z=2\,,\,0\le x\le 1\,$ , so

$$\int\limits_{C_x} F\cdot dr=\int\limits_0^1(4x,2x^2,x^2)\cdot(4,0,0)\,dx=\int\limits_0^116x\,dx=8$$

Try the other two ones.

$$\int\limits_{C_y}F\cdot dr=\int\limits_1^2(4,2,y)\cdot(4,0,0)\,dy=\int\limits_1^216\,dy=16\\{}\\\int\limits_{C_z}F\cdot dr=\int\limits_2^3(2z,4z,2)\cdot(2z,0,0)\,dz=\int\limits_2^34z^2\,dz=\left.\frac{4}{3}z^3\right|_2^3=\frac{4}{3}(27-8)=\frac{76}{3}$$

answered Mar 09 '13 at 01:36

DonAntonio

211,718
17
136
287

for first, it should be (1,0,0) instead of (4,0,0) and thus equal to 2. – sarah Mar 09 '13 at 07:07
for second one again (0,1,0). you are taking derivative of r(t) with respect to t. thus equal to 2 – sarah Mar 09 '13 at 07:15
for third it should be (2t,t,2) and (0,0,1) and integral is equal to 2. – sarah Mar 09 '13 at 07:21
Review line integrals cause you made some mistakes. – sarah Mar 09 '13 at 07:24
Yes, perhaps I messed the limits since I did the above pretty quickly, but it never minds: now you know what to do. – DonAntonio Mar 09 '13 at 09:55

$\int_C ( 2 xz \, \vec {\mathbf i} + x^2z \, \vec {\mathbf j}+ x^2y \, \vec {\mathbf k}) \mathrm d \, \vec {\mathbf r}$ : Line Integral

1 Answers1