Answer to Question #154723 in Calculus for mafizur alam

Question #154723

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


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Expert's answer
2021-01-14T18:55:03-0500

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

SurfaceSurface Integral = 2π12x1+[f(x)]2dx2\pi \int_{1}^{2} x \sqrt{1 + [f'(x)]^2}dx

f(x)=y=4+3x2f(x) = y=4 + 3x^2

f(x)=6xf'(x)=6x


Surface Area = 2π12x1+(6x)2dx2\pi \int_{1}^{2} x \sqrt{1 + (6x)^2}dx


2π12x1+36x2dx2\pi \int_{1}^{2} x \sqrt{1 + 36x^2}dx


let u = 1+36x21 + 36x^2 , dudx=72x{du \over dx} = 72x , dx=du72xdx = {du \over 72x}


Surface Area =2π12x72xudu=2\pi \int_{1}^{2} {x \over 72x} \sqrt{u}du


=2π7212udu={2\pi \over 72} \int_{1}^{2} \sqrt{u}du


=π36.2u3312={\pi \over 36} .{2\sqrt{u}^3 \over 3} |_{1}^{2}

=πu35412= {\pi\sqrt{u}^3 \over54} |_{1}^{2}


=π(1+36x2)35412= {\pi(\sqrt{1 + 36x^2})^3 \over54} |_{1}^{2}


=[π(1+36(2)2)354][π(1+36(1)2)354]=[ {\pi(\sqrt{1 + 36(2)^2})^3 \over54}] - [{\pi(\sqrt{1 + 36(1)^2})^3 \over54} ]

=[π(145)354][π(37)354]=[ {\pi(\sqrt{145})^3 \over54}] - [{\pi(\sqrt{37})^3 \over54} ]



=π54[(145)3(37)3]= {\pi \over54} [ \sqrt{(145)^3}-{\sqrt{(37)^3 }}]

Surface Area =π54[1451453737]= {\pi \over54} [145 \sqrt{145}-37{\sqrt{37 }}]




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