If ∅ = 𝟐𝒙𝒚𝒛 𝟐 , 𝑭 = 𝒙𝒚𝒊 − 𝒚𝒋 + 𝒙 𝟐𝒌 𝐚𝐧𝐝 𝑪 𝐢𝐬 𝐭𝐡𝐞 𝐜𝐮𝐫𝐯𝐞 𝒙 = 𝒕 𝟐 , 𝒚 = 𝟐𝒕, 𝒛 = 𝒕 𝟑 𝐟𝐫𝐨𝐦 𝒕 = 𝟎 𝒕𝒐 𝒕 = 𝟏, 𝐞𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐭𝐡𝐞 𝐥𝐢𝐧𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬 a. ∫ ∅𝒅𝒓 𝑪 , b. ∫ 𝑭 𝑿 𝒅𝒓.
"\\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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