How do you find the surface area of the solid obtained by rotating about the $y$ $y$ -axis the region bounded by $y = 1 - x^{2}$ $y=1-x^2$ on the interval $0 \leq x \leq 1$ $0<=x<=1$ ?

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How do you find the surface area of the solid obtained by rotating about the $y$ $y$ -axis the region bounded by $y = 1 - x^{2}$ $y=1-x^2$ on the interval $0 \leq x \leq 1$ $0<=x<=1$ ?

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Answer 1 · 2014-10-09T18:03:39

The surface area A of the solid obtained by rotating about the $y$ $y$ -axis the region under the graph of $y = f (x)$ $y=f(x)$ from $x = a$ $x=a$ to $b$ $b$ can be found by

$A = 2 π \int_{a}^{b} x \sqrt{1 + {[f (x)]}^{2}} d x$ $A=2pi int_a^b x sqrt{1+[f(x)]^2}dx$ .

Let us now look at the posted question.

By the formula above,

$A = 2 π \int_{0}^{1} x \sqrt{1 + 4 x^{2}} d x$ $A=2pi int_0^1 x sqrt{1+4x^2} dx$

by rewriting a bit,

$= \frac{π}{4} \int_{0}^{1} {(1 + 4 x^{2})}^{\frac{1}{2}} \cdot 8 x d x$ $=pi/4 int_0^1 (1+4x^2)^{1/2}cdot8x dx$

by General Power Rule,

$= \frac{π}{4} {[\frac{2}{3} {(1 + 4 x^{2})}^{\frac{3}{2}}]}_{0}^{\frac{3}{2}} = \frac{π}{6} (5^{\frac{3}{2}} - 1)$ $=pi/4 [2/3(1+4x^2)^{3/2}]_0^1=pi/6(5^{3/2}-1)$

I hope that this was helpful.

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How do you find the surface area of the solid obtained by rotating about the $y$ $y$ -axis the region bounded by $y = 1 - x^{2}$ $y=1-x^2$ on the interval $0 \leq x \leq 1$ $0<=x<=1$ ?

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How do you find the surface area of the solid obtained by rotating about the $y$ $y$ -axis the region bounded by $y = 1 - x^{2}$ $y=1-x^2$ on the interval $0 \leq x \leq 1$ $0<=x<=1$ ?