What is the surface area of the solid created by revolving $f (x) = 3 x, x \in [2, 5]$ $f(x) = 3x, x in [2,5]$ around the x axis?

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Answer 1 · 2016-05-24T10:17:17

$A = 63 \sqrt{10} π$ $A=63sqrt10 pi$

$A = 2 π \int_{2}^{5} f (x) \cdot {\sqrt{1 + \frac{d f (x)}{d x}}}^{2} \cdot d x$ $A=2pi int_2^5 f(x)*sqrt(1+(d f(x))/(d x))^2*d x$

$f (x) = 3 x$ $f(x)=3x$

$\frac{d}{d x} f (x) = 3$ $d/(d x) f(x)=3$

$A = 2 π \int_{2}^{5} 3 x \cdot \sqrt{1 + 3^{2}} \cdot d x$ $A=2 pi int_2^5 3x*sqrt(1+3^2)* d x$

$A = 2 π \int_{2}^{5} 3 x \sqrt{10} \cdot d x$ $A=2 pi int_2^5 3xsqrt 10 *d x$

$A = 6 \sqrt{10} π \int_{2}^{5} x \cdot d x$ $A=6sqrt10 pi int_2^5 x *d x$

$A = 6 \sqrt{10} {[\frac{1}{2} x^{2}]}_{2}^{2}$ $A=6 sqrt10[1/2 x^2]_2^5$

$A = 6 \sqrt{10} π [(\frac{1}{2} \cdot 5^{2} - \frac{1}{2} \cdot 2^{2})]$ $A=6sqrt10 pi[(1/2*5^2-1/2*2^2)]$

$A = 6 \sqrt{10} π [\frac{25}{2} - \frac{4}{2}]$ $A=6sqrt 10 pi[25/2-4/2]$

$A = 6 \sqrt{10} π [\frac{21}{2}]$ $A=6sqrt 10 pi[21/2]$

$A = 63 \sqrt{10} π$ $A=63sqrt10 pi$

What is the surface area of the solid created by revolving f(x)=3x,x∈[2,5]f(x) = 3x, x in [2,5] around the x axis?