Question

If F=A×(B×C) where A=5it^2 + (3t-2)j + 6tk; B=2i+4tj+3(1-t)k and C=4ti+5jt^2 -3tk evaluate integral of Fdt with limits zero to one.

Answer 1

$$\vec{A}=5t^2\vec{i}+(3t-2)\vec{j}+6t\vec{k}$$ $$\vec{B}=2\vec{i}+4t\vec{j}+3(1-t)\vec{k}$$ $$\vec{C}=4t\vec{i}+5t^2\vec{j}-3t\vec{k}$$

in a right hand coordinate system:

$$\vec{D}=\vec{B}\times\vec{C}=(B_yC_z-B_zC_y)\vec{i}+(B_zC_x-B_xC_z)\vec{j}+(B_xC_y-B_yC_x)\vec{k}=$$ $$(15t^3-25t^2)\vec{i}+(18t-12t^2)\vec{j}-6t^2\vec{k}$$ $$\vec{F}=\vec{A}\times(\vec{B}\times\vec{C})=\vec{A}\times\vec{D}=$$ $$(A_yD_z-A_zD_y)\vec{i}+(A_zD_x-A_xD_z)\vec{j}+(A_xD_y-A_yD_x)\vec{k}=$$ $$(54t^3-96t^2)\vec{i}+(120t^4-162t^3)\vec{j}+(201t^3-105t^4-54t^2)\vec{k}$$

$$\underset{0}{\overset{1}{\int}}\vec{F}dt=$$ $$\underset{0}{\overset{1}{\int}}((54t^3-96t^2)\vec{i}+(120t^4-162t^3)\vec{j}+(201t^3-105t^4-54t^2)\vec{k})dt=$$ $$\vec{i}\underset{0}{\overset{1}{\int}}(54t^3-96t^2)dt+\vec{j}\underset{0}{\overset{1}{\int}}(120t^4-162t^3)dt+\vec{k}\underset{0}{\overset{1}{\int}}(201t^3-105t^4-54t^2)dt=$$ $$\vec{i}(\frac{27t^4}{2}-32t^3)\mid_{0}^1+\vec{j}(24t^5-\frac{81t^4}{2})\mid_{0}^1+\vec{k}(\frac{201t^4}{4}-21t^5-18t^3)\mid_{0}^1=$$ $$-\frac{37}{2}\vec{i}-\frac{33}{2}\vec{j}+\frac{45}{4}\vec{k}$$

Answer:

$\underset{0}{\overset{1}{\int}}\vec{F}dt=-\frac{37}{2}\vec{i}-\frac{33}{2}\vec{j}+\frac{45}{4}\vec{k}$