Question #184946

Determine the surface area generated by revolving the curve y=1/3 x³ over the interval (0,2), about the x-axis.

Expert's answer

**Solution**

If the curve y = f(x), a≤x≤b is rotated about the x-axis, then the surface area is given by

"A=2\\pi\\int_a^bf(x)\\sqrt{1+[f'(x)]^2}dx"

In this case f(x) = 1/3 x³, f’(x) = x^{2} ,a = 0, b = 2

"A=2\\pi\\int_0^2\\frac{1}{3}x^3\\sqrt{1+x^4}dx"

Substitution y = x^{4}

"A=\\frac{\\pi}{6}\\int_0^{16}\\sqrt{1+y}dy=\\frac{\\pi}{9}(1+y)^{3\/2}|_0^{16}=\\frac{\\pi}{9}(17^{3\/2}-1)=24.118"

**Answer**

"A=\\frac{\\pi}{9}(17^{3\/2}-1)=24.118"

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