anonymous
  • anonymous
Evaluate the line integral with respect to arc length (intergral)(3*(x^2)*y*z)ds Where c is given by X=t, y=t^2, z=(2/3)*(t^3), 0<=t<=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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watchmath
  • watchmath
\(\int_0^1 3t^2\cdot t^2\cdot \frac{2}{3}t^3\sqrt{1^2+(2t)^2+(2t^2)^2}\,dt\) \(\int_0^12t^7\sqrt{((2t^2)+1)^2}\,dt\) \(\int_0^1 2t^7\cdot ( (2t)^2+1)\, dt\)
anonymous
  • anonymous
thanks mate, do you reckon you could explain how you got from \[\sqrt{1^{2}+2t^2+(2t^2)^2}\] to \[\sqrt{((2t^2)+1)^2 }\]
watchmath
  • watchmath
no, i didn't say that :) I said that \(1^2+(2t)^2+(2t^2)^2=4t^4+4t^2+1=((2t^2)+1)^2\) You can easily check that if you expand \(((2t^2)+1)^2\).

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