What is the surface area of the solid created by revolving $f (x) = x \sqrt{x - 1}$ $f(x)=xsqrt(x-1)$ for $x \in [2, 3]$ $x in [2,3]$ around the x-axis?

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What is the surface area of the solid created by revolving $f (x) = x \sqrt{x - 1}$ $f(x)=xsqrt(x-1)$ for $x \in [2, 3]$ $x in [2,3]$ around the x-axis?

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Answer 1 · 2018-03-12T14:16:42

Area $\approx 19.363$ $~~ 19.363$ square units (to be specified from the units of $x$ $x$ )

Explanation:

Double integration may be applied with the element of summation comprising some notional approximate rectangle (becoming exact in the limit as the length of the sides approaches $0$ $0$ ) of sides $d x$ $dx$ and $r d θ$ $r d theta$ where $r$ $r$ is the radius of the circle being swept to form the surface, such that, in this case, $r = x \sqrt{x - 1}$ $r = x sqrt(x - 1)$ and $θ$ $theta$ is the angle through which the radius is swept. It will be necessary to sweep from $θ = 0$ $theta = 0$ to $θ = 2 π$ $theta = 2 pi$ , over the interval $x = 2$ $x = 2$ to $x = 3$ $x = 3$ .

Denoting the required area by A, retaining $r$ $r$ for the moment to illustrate the point, the integral may be set up as

$A = \int_{θ = 0}^{θ = 2 π} \int_{x = 2}^{x = 3} r d x d θ$ $A = int_(theta = 0)^(theta = 2pi) int_(x = 2)^(x = 3) r dx d theta$

that is

$A = \int_{θ = 0}^{θ = 2 π} \int_{x = 2}^{x = 3} x \sqrt{x - 1} d x d θ$ $A = int_(theta = 0)^(theta = 2pi) int_(x = 2)^(x = 3) xsqrt(x-1) dx d theta$

Taking the inner integral first,

$\int_{2}^{3} x \sqrt{x - 1} d x$ $int_2^3 xsqrt(x-1) dx$

this may be solved using the substitution

$u (x) = x - 1$ $u(x) = x - 1$

so that

$\frac{d u}{d x} = 1$ $(du)/dx = 1$

so that

$\int d u = \int d x$ $int du = int dx$

Noting
$u = x - 1$ $u = x - 1$
implies
$x = 1 + u$ $x = 1 + u$
so that
$x \sqrt{x - 1} = (1 + u) \sqrt{u} = \sqrt{u} + u^{\frac{3}{2}}$ $xsqrt(x-1) = (1 + u) sqrt(u) = sqrt(u) + u^(3/2)$

and, for the limits,
u(2) = 1
and
u(3) = 2

the required (inner) integral is

$\int_{1}^{2} (u^{\frac{1}{2}} + u^{\frac{3}{2}}) d u$ $int_1^2 (u^(1/2) + u^(3/2)) du$

which, by the sum rule is

$\int_{1}^{2} u^{\frac{1}{2}} d u + \int_{1}^{2} u^{\frac{3}{2}} d u$ $int_1^2 u^(1/2) du + int_1^2 u^(3/2) du$

$= (\frac{2}{3}) {[u^{\frac{3}{2}}]}_{1}^{\frac{3}{2}} + (\frac{2}{5}) {[u^{\frac{5}{2}}]}_{1}^{\frac{5}{2}}$ $= (2/3)[u^(3/2)]_1^2 + (2/5)[u^(5/2)]_1^2$

$= (\frac{2}{3}) (2^{\frac{3}{2}} - 1) + (\frac{2}{5}) (2^{\frac{5}{2}} - 1)$ $= (2/3)(2^(3/2)-1) + (2/5)(2^(5/2) -1)$

$\approx 3.0817$ $~~ 3.0817$ (to 4 decimal places, courtesy of Gnu Octave but defer that collapsing to an approximate value until the outer integral is evaluated ... )

Denoting the first (inner) integral by $K$ $K$ (as it evaluates to a constant), the outer integral may now be evaluated

$A = \int_{0}^{2 π} K d θ$ $A = int_0^(2pi) K d theta$

$= K {[θ]}_{0}^{2 π}$ $= K[theta]_0^(2pi)$

$= 2 π K$ $= 2 pi K$

So the overall double integral evaluates to

$A = 2 π ((\frac{2}{3}) (2^{\frac{3}{2}} - 1) + (\frac{2}{5}) (2^{\frac{5}{2}} - 1))$ $A = 2 pi ((2/3)(2^(3/2)-1) + (2/5)(2^(5/2) -1))$

$\approx 19.363$ $~~19.363$ square units (to be specified from the units of $x$ $x$ )
(to 3 decimal places, again courtesy of Gnu Octave)

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What is the surface area of the solid created by revolving $f (x) = x \sqrt{x - 1}$ $f(x)=xsqrt(x-1)$ for $x \in [2, 3]$ $x in [2,3]$ around the x-axis?

1 Answer

Explanation:

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What is the surface area of the solid created by revolving f(x)=x√x−1f(x)=xsqrt(x-1) for x∈[2,3]x in [2,3] around the x-axis?

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What is the surface area of the solid created by revolving $f (x) = x \sqrt{x - 1}$ $f(x)=xsqrt(x-1)$ for $x \in [2, 3]$ $x in [2,3]$ around the x-axis?