What is the surface area of the solid created by revolving $f (x) = 2 - x$ $f(x)=2-x$ over $x \in [2, 3]$ $x in [2,3]$ around the x-axis?

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Answer 1 · 2016-04-17T13:22:10

$A = - \sqrt{2} π$ $A=-sqrt2 pi$

$A = 2 π \int_{2}^{3} f (x) \sqrt{1 + {(\frac{d}{d x} f (x))}^{2}} d x$ $A=2piint_2^3 f(x)sqrt(1+(d/(d x) f(x))^2 )d x$

$\frac{d}{d x} f (x) = - 1$ $d/(d x)f(x)=-1$

${(\frac{d}{d x} f (x))}^{2} = 1$ $(d/(d x)f(x))^2=1$

$A = 2 π \int_{2}^{3} (2 - x) \sqrt{1 + 1} d x$ $A=2pi int_2^3 (2-x)sqrt(1+1)d x$

$A = 2 π \int_{2}^{3} (2 - x) \sqrt{2} \cdot d x$ $A=2 pi int _2^3 (2-x)sqrt2* d x$

$A = 2 \sqrt{2} π \int_{2}^{3} (2 - x) d x$ $A=2sqrt2 pi int_2^3 (2-x)d x$

$A = 2 \sqrt{2} π {∣ ∣ ∣ 2 x - \frac{x^{2}}{2} ∣ ∣ ∣}_{2}^{3}$ $A=2 sqrt2 pi|2x-x^2/2|_2^3$

$A = 2 \sqrt{2} π [(2 \cdot 3 - \frac{3^{2}}{2}) - (2 \cdot 2 - \frac{2^{2}}{2})]$ $A=2 sqrt2 pi[(2*3-3^2/2)-(2*2-2^2/2)]$

$A = 2 \sqrt{2} π [(6 - \frac{9}{2}) - (4 - 2)]$ $A=2 sqrt2 pi[(6-9/2)-(4-2)]$

$A = 2 \sqrt{2} π [\frac{3}{2} - 2]$ $A=2sqrt2 pi[3/2-2]$

$A = 2 \sqrt{2} π (- \frac{1}{2})$ $A=2sqrt2 pi(-1/2)$

$A = - \sqrt{2} π$ $A=-sqrt2 pi$

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