Answer to Question #201072 in Calculus for Rocky Valmores

Question #201072

If โˆ… = ๐Ÿ๐’™๐’š๐’› ๐Ÿ , ๐‘ญ = ๐’™๐’š๐’Š โˆ’ ๐’š๐’‹ + ๐’™ ๐Ÿ๐’Œ ๐š๐ง๐ ๐‘ช ๐ข๐ฌ ๐ญ๐ก๐ž ๐œ๐ฎ๐ซ๐ฏ๐ž ๐’™ = ๐’• ๐Ÿ , ๐’š = ๐Ÿ๐’•, ๐’› = ๐’• ๐Ÿ‘ ๐Ÿ๐ซ๐จ๐ฆ ๐’• = ๐ŸŽ ๐’•๐’ ๐’• = ๐Ÿ, ๐ž๐ฏ๐š๐ฅ๐ฎ๐š๐ญ๐ž ๐ญ๐ก๐ž ๐ฅ๐ข๐ง๐ž ๐ข๐ง๐ญ๐ž๐ ๐ซ๐š๐ฅ๐ฌ a. โˆซ โˆ…๐’…๐’“ ๐‘ช , b. โˆซ ๐‘ญ ๐‘ฟ ๐’…๐’“.


1
Expert's answer
2021-07-01T17:46:00-0400

"\\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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