Answer to Question #155139 in Calculus for haana

Question

Evaluate the integrals

i) $\int_{s}^{}\overrightarrow{F} \overrightarrow{ds}$

ii) $\int_{s}^{}\overrightarrow{F} \times \overrightarrow{ds}$

$\overrightarrow{F} = ( x^2 +y , y + z , z + x)$ and s is the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑧 ≤ 2 and 𝑦 = 0 positively oriented along the positive Y-axis.

Answer 1

for this area $\vec{ds}=\vec{j}ds=\vec{j}dxdz$

(i)

$\int_S\vec{F}\vec{ds}=\int_0^2\int_0^1((x^2+y)\vec{i}+(y+z)\vec{j}+(z+x)\vec{k})\vec{j}dxdz=$

$=\int_0^2\int_0^1(y+z)dxdz=\int_0^2\int_0^1zdxdz$=

(y=0 on this area)

$=\int_0^2zx|_{x=0}^{x=1}dz=\int_0^2zdz=\frac{z^2}{2}|_0^2=2$

(ii)

$\int_S\vec{F}\times\vec{ds}=$

$=\int_S\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ F_x & F_y & F_z \\ ds_x &ds_y & ds_z \end{vmatrix}=$

(here we multiply with $ds_x=0$ and $ds_z=0$ )

$=\int_S(\vec{i}(-F_zds_y)+\vec{k}F_xds_y)=$

$=\int_0^2\int_0^1(-\vec{i}(z+x)+\vec{k}(x^2+y))dxdz=$

(y=0 on this area)

$=\int_0^2\int_0^1(-\vec{i}(z+x)+\vec{k}x^2)dxdz=$

$=\int_0^2(-\vec{i}(zx+\frac{x^2}{2})|_{x=0}^{x=1}+\vec{k}\frac{x^3}{3}|_{x=0}^{x=1})dz=$

$=\int_0^2(-\vec{i}(z+\frac{1}{2})+\vec{k}\frac{1}{3})dz=$

$=-\vec{i}(\frac{z^2}{2}+\frac{z}{2})|_0^2+\vec{k}\frac{z}{3}|_0^2=$

$=-3\vec{i}+\frac{2}{3}\vec{k}$