Answer to Question #154723 in Calculus for mafizur alam

Question

FIND THE SURFACE AREA OF THE OBJECT OBTAINED BY ROTATING Y=4+3x^2, 1<=x<=2 ABOUT THE Y-AXIS

Answer 1

FIND THE SURFACE AREA OF THE OBJECT OBTAINED BY ROTATING Y=4+3x², 1<=x<=2 ABOUT THE Y-AXIS

$Surface$ Integral = $2\pi \int_{1}^{2} x \sqrt{1 + [f'(x)]^2}dx$

$f(x) = y=4 + 3x^2$

$f'(x)=6x$

Surface Area = $2\pi \int_{1}^{2} x \sqrt{1 + (6x)^2}dx$

$2\pi \int_{1}^{2} x \sqrt{1 + 36x^2}dx$

let u = $1 + 36x^2$ , ${du \over dx} = 72x$ , $dx = {du \over 72x}$

Surface Area $=2\pi \int_{1}^{2} {x \over 72x} \sqrt{u}du$

$={2\pi \over 72} \int_{1}^{2} \sqrt{u}du$

$={\pi \over 36} .{2\sqrt{u}^3 \over 3} |_{1}^{2}$

$= {\pi\sqrt{u}^3 \over54} |_{1}^{2}$

$= {\pi(\sqrt{1 + 36x^2})^3 \over54} |_{1}^{2}$

$=[ {\pi(\sqrt{1 + 36(2)^2})^3 \over54}] - [{\pi(\sqrt{1 + 36(1)^2})^3 \over54} ]$

$=[ {\pi(\sqrt{145})^3 \over54}] - [{\pi(\sqrt{37})^3 \over54} ]$

$= {\pi \over54} [ \sqrt{(145)^3}-{\sqrt{(37)^3 }}]$

Surface Area $= {\pi \over54} [145 \sqrt{145}-37{\sqrt{37 }}]$