What is the surface area of the solid created by revolving $f (x) = \frac{e^{x}}{2}, x \in [2, 7]$ $f(x) = e^(x)/2 , x in [2,7]$ around the x axis?

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What is the surface area of the solid created by revolving $f (x) = \frac{e^{x}}{2}, x \in [2, 7]$ $f(x) = e^(x)/2 , x in [2,7]$ around the x axis?

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Answer 1 · 2016-11-03T15:54:49

The area is $= \frac{π}{8} ((4 (ln (e^{14} + 4) + e^{7}) + e^{7} \sqrt{e^{14} + 4})) - \frac{π}{8} (4 (ln (e^{4} + 4) + e^{2}) + e^{2} \sqrt{e^{4} + 4}))$ $=pi/8((4(ln(e^(14)+4)+e^7)+e^7sqrt(e^(14)+4)))-pi/8(4(ln(e^(4)+4)+e^2)+e^2sqrt(e^(4)+4)))$

Explanation:

The volume of a small disc is $d S = π y d r$ $dS=piydr$
and $d r = \sqrt{{(d x)}^{2} + {(d y)}^{2}}$ $dr=sqrt((dx)^2+(dy)^2)$
$d S = π y \sqrt{{(d x)}^{2} + {(d y)}^{2}} = π y \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $dS=piysqrt((dx)^2+(dy)^2)=piysqrt(1+(dy/dx)^2)dx$
So $y = \frac{e^{x}}{2}$ $y=e^x/2$ $\Rightarrow$ $=>$ $\frac{d y}{d x} = \frac{e^{x}}{2}$ $dy/dx=e^x/2$
so $d S = π \frac{e^{x}}{2} \cdot \sqrt{1 + \frac{e^{2 x}}{4}} d x$ $dS=pie^x/2*sqrt(1+e^(2x)/4)dx$
$d S = π \frac{e^{x}}{4} \cdot \sqrt{4 + e^{2 x}} d x$ $dS=pie^x/4*sqrt(4+e^(2x))dx$
so $S = \frac{π}{4} \int e^{x} \sqrt{4 + e^{2 x}} d x$ $S=pi/4inte^xsqrt(4+e^(2x))dx$
let $u = e^{x}$ $u=e^x$ $\Rightarrow$ $=>$ $d u = e^{x} d x$ $du=e^xdx$
$S = \frac{π}{4} \cdot \int \sqrt{4 + u^{2}} d u$ $S=pi/4*intsqrt(4+u^2)du$
$u = 2 tan v$ $u=2tanv$ $\Rightarrow$ $=>$ $d u = {sec}^{2} u (d v)$ $du= sec^2u (dv)$
and $4 {tan}^{2} v + 4 = 4 {sec}^{2} v$ $4tan^2v+4=4sec^2v$
$S = 4 \frac{π}{4} \int {sec}^{3} v d v = π (\frac{1}{2} \int sec v d v + sec v \frac{tan v}{2})$ $S=4pi/4intsec^3vdv=pi(1/2intsecvdv+secvtanv/2)$
$π ((\frac{ln (tan v + sec v)}{2}) + sec v \frac{tan v}{2})$ $pi((ln(tanv+secv)/2)+secvtanv/2)$
$S = \frac{π}{8} (4 (ln (e^{2 x} + 4) + e^{x}) + e^{x} \sqrt{e^{2 x} + 4}))_{2}^{7}$ $S=pi/8(4(ln(e^(2x)+4)+e^x)+e^xsqrt(e^(2x)+4)))_2^7$
$= \frac{π}{8} ((4 (ln (e^{14} + 4) + e^{7}) + e^{7} \sqrt{e^{14} + 4})) - \frac{π}{8} (4 (ln (e^{4} + 4) + e^{2}) + e^{2} \sqrt{e^{4} + 4}))$ $=pi/8((4(ln(e^(14)+4)+e^7)+e^7sqrt(e^(14)+4)))-pi/8(4(ln(e^(4)+4)+e^2)+e^2sqrt(e^(4)+4)))$

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What is the surface area of the solid created by revolving $f (x) = \frac{e^{x}}{2}, x \in [2, 7]$ $f(x) = e^(x)/2 , x in [2,7]$ around the x axis?

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Explanation:

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What is the surface area of the solid created by revolving f(x)=ex2,x∈[2,7]f(x) = e^(x)/2 , x in [2,7] around the x axis?

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What is the surface area of the solid created by revolving $f (x) = \frac{e^{x}}{2}, x \in [2, 7]$ $f(x) = e^(x)/2 , x in [2,7]$ around the x axis?