Answer to Question #154224 in Calculus for mafizur alam

Question

find the surface area of the object obtained by rotating y=4+3x^2, 1<=x<=2

Answer 1

Surface area obtained by rotating the curve about y axis

will be

S=$\int_{1}^{2}2\pi x\sqrt{(1+(y_1)^2)}dx$ [ where the $y_1$ is the differentiation with respect to x]

$\implies\int_1^22\pi x\sqrt{(1+(6x)^2)}dx\\\implies\frac{\pi}{36}\int_1^272 x\sqrt{(1+(6x)^2)}dx\\\implies\frac{\pi}{36}\int_{36}^{144}\sqrt{(1+z)}dz\\\implies\frac{\pi}{36}\frac{2}{3}(1+z)^\frac{3}{2}|_{36}^{144}\\\implies\frac{\pi}{54}[145^\frac{3}{2}-37^\frac{3}{2}]$ [Here we take Z=36x²]

So area of the given curve is $\frac{\pi}{54}[145^\frac{3}{2}-37^\frac{3}{2}]$