Question
Asked Nov 6, 2019
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find the exact area of the surface obtained by rotating x=t^3,y=t^2, 0<t<1 about the x-axis.

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Expert Answer

Step 1

y=t^2 so t= sqrt(y)

x=t^3= (sqrt(y))^3 = y^(3/2)

x=y^(3/2) (graphed this equation)

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1 0.5 t 2 -05 0.5 15 1.5 AXIS OF ROTATION 0.5

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Step 2

Use the surface area formula .

There plug x=t^3 and y=t^2 then simplify.

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The surface area is S 2Ty ds 2 'dy da 2nt2 dt dt dt 2 (d(t3) (d (2) 2Tt2 dt dt dt 2nt2/(32 + (20)? dt 2mt2 /9¢* + 4? dt t3 912+4 dt - 2 T

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Step 3

When t=0 , then u=4 

When t=1 then u=1...

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We will substitute 942 +4 = u du And 18t dt = du t dt 18 Also u4 t2 9 e13 Limits of Integration will change from to

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Math

Calculus