If โ = ๐๐๐๐ ๐ , ๐ญ = ๐๐๐ โ ๐๐ + ๐ ๐๐ ๐๐ง๐ ๐ช ๐ข๐ฌ ๐ญ๐ก๐ ๐๐ฎ๐ซ๐ฏ๐ ๐ = ๐ ๐ , ๐ = ๐๐, ๐ = ๐ ๐ ๐๐ซ๐จ๐ฆ ๐ = ๐ ๐๐ ๐ = ๐, ๐๐ฏ๐๐ฅ๐ฎ๐๐ญ๐ ๐ญ๐ก๐ ๐ฅ๐ข๐ง๐ ๐ข๐ง๐ญ๐๐ ๐ซ๐๐ฅ๐ฌ a. โซ โ ๐ ๐ ๐ช , b. โซ ๐ญ ๐ฟ ๐ ๐.
"\\phi=2xyz^{2}"
"F=xy""\\hat{i}-y\\hat{j}+x^{2}\\hat{k}"
Curve C is given by "x=t^{2},y=2t,z=t^{3}" from t=0 to t=1 .
We have to evaluate line integeral -
"a)\\int_{c}\\phi" "dr"
"d\\vec{r}=dx\\hat{i}+dy\\hat{j}+dz\\hat{k}"
"\\phi" ="2xyz^{2}" , now making this variable in terms of t , we have to put the given "x,y\\ and \\ z" so that we can get a function in t .
"x=t^{2}" ",y=2t" ",z=t^{3}"
"\\phi=" "2" "t^{2}\\times2t\\times(t^{3})^{2}"
"\\phi=4t^{9}"
"dx=2t""dt" "," "dy=2dt" ",dz=3t^{2}dt"
"d\\hat{r}=(2t\\hat{i}+2\\hat{j}+3t^{2}\\hat{k})dt"
"=\\int_c\\phi" "d\\hat{r}=\\int_{0}^{1}(" "2t\\hat{i}+2\\hat{j}+3t^{2}\\hat{k})" ".4t^{9}dt"
"=\\int_0^1(8t^{10}\\hat{i}+8t^{9}\\hat{j}+12t^{11}\\hat{k})dt"
"=\\dfrac{8}{11}\\hat{i}+\\dfrac{8}{10}\\hat{j}+t^{12}\\hat{k}"
"B)" "\\int F.Xdr"
"d\\hat{r}=dx\\hat{i}+dy\\hat{j}+dz\\hat{k}"
"d\\hat{r}=(2t\\hat{i}+2\\hat{j}" "+3t^{2}\\hat{k})dt"
"F=xy\\hat{i}-y\\hat{j}+x^{2}\\hat{k}"
now putting the value of "x, y ,z" in above equation we get -
"F=" "2t^{3}\\hat{i}-2t\\hat{j}+t^{4}\\hat{k}"
"=\\int_{0}^{1}(2t^{3}\\hat{i}-2t\\hat{j}+t^{4}\\hat{k}).(2t\\hat{i}+2\\hat{j}+3t^{2}\\hat{k})dt"
"=\\int_{0}^{1}(4t^{4}-4t+3t^{6})dt"
"=\\dfrac{4}{5}-2+\\dfrac{3}{7}"
"=\\dfrac{28-70+15}{35}=\\dfrac{-27}{35}"
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