What is the surface area of the solid created by revolving $f (x) = \sqrt{x}$ $f(x)=sqrt(x)$ for $x \in [1, 2]$ $x in [1,2]$ around the x-axis?

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What is the surface area of the solid created by revolving $f (x) = \sqrt{x}$ $f(x)=sqrt(x)$ for $x \in [1, 2]$ $x in [1,2]$ around the x-axis?

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Answer 1 · 2018-05-11T23:53:02

$\frac{π}{6} [27 - 5 \sqrt{5}]$ $pi/6[27-5sqrt5]$ units^ $2$ $2$

Explanation:

The formula for surface area of revolution is, $S A = 2 π \int_{a}^{b} f [x] \sqrt{1 + {[\frac{d}{d x} f [x]]}^{2} d x}$ $SA=2piint_a^bf[x]sqrt[1+[d/dxf[x]]^2dx$ .

From the question, $S A$ $SA$ = $2 π \int_{1}^{2} \sqrt{x} \sqrt{1 + {[\frac{d}{d x} \sqrt{x}]}^{2}}$ $2piint_1^2sqrtxsqrt[1+[d/dxsqrtx]^2$ [where $f [x]$ $f[x]$ = $\sqrt{x}]$ $sqrtx]$ .

Differentiating $\frac{d}{d x} \sqrt{x}$ $d/dxsqrtx$ = $\frac{1}{2 \sqrt{x}}$ $1/[2sqrtx$ and so ${[\frac{d}{d x} [\sqrt{x}]]}^{2}$ $[d/dx[sqrtx]]^2$ = $\frac{1}{4 x}$ $1/[4x$

Therefore, $S A$ $SA$ = $2 π \int_{1}^{2} \sqrt{x} \sqrt{1 + \frac{1}{4 x} d x}$ $2piint_1^2sqrtxsqrt[1+1/[4x]dx$ = $2 π \int_{1}^{2} \sqrt{x} \sqrt{\frac{4 x + 1}{4 x} d x}$ $2piint_1^2sqrtxsqrt[[4x+1]/[4x]dx$ .

That is, $2 π \int_{1}^{2} \frac{\sqrt{x}}{2 \sqrt{x}}$ $2piint _1^2sqrtx/[2sqrtx]$ $\sqrt{4 x + 1}$ $sqrt[4x+1$ dx = $π \int_{1}^{2} \sqrt{4 x + 1}$ $piint_1^2sqrt[4x+1$ dx.......... $[1]$ $[1]$

Integrating .... $[1]$ $[1]$ w.r.t. gives us $S A$ $SA$ =, $\frac{π}{6} \sqrt{{[4 x + 1]}^{3}}$ $pi/6sqrt[[4x+1]^3]$ and when evaluted for $x = 1, x = 2$ $x= 1, x=2$ will result inthe above answer when simplified.

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What is the surface area of the solid created by revolving $f (x) = \sqrt{x}$ $f(x)=sqrt(x)$ for $x \in [1, 2]$ $x in [1,2]$ around the x-axis?

1 Answer

Explanation:

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What is the surface area of the solid created by revolving f(x)=sqrt(x)f(x)=√xf(x)=sqrt(x) for x in [1,2]x∈[1,2]x in [1,2] around the x-axis?

1 Answer

Explanation:

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What is the surface area of the solid created by revolving $f (x) = \sqrt{x}$ $f(x)=sqrt(x)$ for $x \in [1, 2]$ $x in [1,2]$ around the x-axis?