Test 1 Solutions-2018SP-MA Wi 10.3 Chegg Study | Guided Solutior x www.webassign

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Test 1 Solutions-2018SP-MA Wi 10.3 Chegg Study | Guided Solutior x www.webassign.net/web/Student/Assignment-Responses/last?dep-18577942 13 points LarCalc10 10.3.062. Write an integral that represents the area of the surface generated by revolving the curve about the x-axis. Parametric Equations Interval dt Use a graphing utility to approximate the integral. (Round your answer to three decimal places.) Need Help? Read It Talk toa Tutor 5. O-12 points LarCalc10 10.3.065 Find the area of the surface generated by revolving the curve about each given axis (a) x-axis

Explanation / Answer

We are given the curve in the parametric form that is:

x and y are given in term of a parameter 't' and t E [a,b] and this curve revolves around the x-axis.

For such problems the area of the solid so formed is given by the integral :

Area , A = 2pi*Integral(t=a to b) [y(t)sqrt[(dx/dt)^2 + (dy/dt)^2]]dt

Now we are given : x(t) = 1/4*t^2 and y(t) = 8t + 2 and t E [0 , 9]

=> dx/dt = t/2 , (dx/dt)^2 = 1/4*t^2

and dy/dt = 8 , (dy/dt)^2 = 64

and a = 0 and b = 9

=> Area , A = 2pi*Integral(t=a to b) [y(t)sqrt[(dx/dt)^2 + (dy/dt)^2]]dt

=> Area , A = 2pi*Integral(t=0 to 9) [(8t + 2)sqrt[1/4*t^2 + 64]]dt ----> This is the required integral

Now lets use wolframalpha to approximate the value of the integral :

=> The area is = 18463.762 (approximate value)

  

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