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

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Answer 1 · 2014-10-09T18:03:39

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

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

Let us now look at the posted question.

By the formula above,

$A=2pi int_0^1 x sqrt{1+4x^2} dx$

by rewriting a bit,

$=pi/4 int_0^1 (1+4x^2)^{1/2}cdot8x dx$

by General Power Rule,

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

I hope that this was helpful.

How do you find the surface area of the solid obtained by rotating about the yy-axis the region bounded by y=1-x^2y=1-x^2 on the interval 0<=x<=10<=x<=1 ?