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

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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 $y = x^{2}$ $y=x^2$ on the interval $1 \leq x \leq 2$ $1<=x<=2$ ?

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Answer 1 · 2014-09-05T02:10:59

First of all, you are missing a bound. We will assume that the other bound is $y = 0$ $y=0$ or the $x$ $x$ -axis. The answer is $\frac{15 π}{2}$ $(15pi)/2$ .

The first step is to determine whether you are rotating along an axis that is parallel to the independent axis or the axis of the parameter ( $x$ $x$ in this case). And we are not, so this integration should be done with cylindrical shells.

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Always draw a diagram to verify what is the parameter and what is the function.

You should note that $y$ $y$ is not always a parameter of $x$ $x$ . For instance, $x = y^{2}$ $x=y^2$ , $x$ $x$ is now a parameter of $y$ $y$ .

The formula for cylindrical shells is:

$V = \int_{a}^{b} 2 π r h d r$ $V=int_a^b2pirhdr$
$h$ $h$ is represented by $y$ $y$ , we have $y = x^{2}$ $y=x^2$ and $y = 0$ $y=0$
$r$ $r$ is represented by $x$ $x$
$V = \int_{1}^{2} 2 π x (x^{2} - 0) d x$ $V=int_1^2 2 pi x (x^2-0) dx$

Now that the substitutions are done, we can solve:

$V = 2 π \int_{1}^{2} x^{3} d x$ $V=2 pi int_1^2x^3dx$
$= 2 π \frac{x^{4}}{4} {∣ ∣ ∣}_{1}^{2}$ $=2pi (x^4)/4|_1^2$
$= 2 π \frac{[2^{4} - 1^{4}]}{4}$ $=2pi([2^4-1^4])/4$
$= \frac{15 π}{2}$ $=(15pi)/2$

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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 $y = x^{2}$ $y=x^2$ on the interval $1 \leq x \leq 2$ $1<=x<=2$ ?

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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 $y = x^{2}$ $y=x^2$ on the interval $1 \leq x \leq 2$ $1<=x<=2$ ?