Set up an integral for the area of the surface generated by revolving the curve y...

 

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Set up an integral for the area of the surface generated by revolving the curve y +4√y = x over 8 ≤ y ≤ 14 about the y-axis. Try graphing the curve and drawing a picture of what the surface looks like. Surface Area = b dy where a = and b = Find the area of the surface generated by revolving the curve y = √√x +3 over 1 ≤ x ≤ 8 about the x-axis. Surface area is Find the area of the surface generated by revolving the curve x = 1/²2 3 Surface area is over 0 ≤ y ≤ 1 about the y-axis. JA Find the area of the surface generated by revolving the curve x = axis. 1 8y² Now evaluate the integral S = [2лyds with appropriate limits. Surface area is from 1 ≤ y ≤ 3 about the x- Hint: Express ds = √dx² + dy² in terms of dy. Enter that expression here: ds = Set up an integral for the area of the surface generated by revolving the curve y +4√y = x over 8 ≤ y ≤ 14 about the y-axis. Try graphing the curve and drawing a picture of what the surface looks like. Surface Area = b dy where a = and b = Find the area of the surface generated by revolving the curve y = √√x +3 over 1 ≤ x ≤ 8 about the x-axis. Surface area is Find the area of the surface generated by revolving the curve x = 1/²2 3 Surface area is over 0 ≤ y ≤ 1 about the y-axis. JA Find the area of the surface generated by revolving the curve x = axis. 1 8y² Now evaluate the integral S = [2лyds with appropriate limits. Surface area is from 1 ≤ y ≤ 3 about the x- Hint: Express ds = √dx² + dy² in terms of dy. Enter that expression here: ds =

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