# What is the surface area produced by rotating f(x)=1-x, x in [0,3] around the x-axis?

Mar 1, 2017

$S A = 5 \sqrt{2} \pi$

#### Explanation:

graph{(y-1+x)=0 [-10, 10, -5, 5]}

graph{(y-1+x)(y+1-x)=0 [-10, 10, -5, 5]}

Method 1

The surface area will be that of the combination of two cones;
- smaller cone; h=1; r=1; l=sqrt(2)
- larger cone; h=2; r=2; l=2sqrt(2)

The surface area of a cone is given by $S A = \pi r l$, ($l$=slant)

Hence:

$S A = \pi \left(1\right) \sqrt{2} + \pi \left(2\right) 2 \sqrt{2}$
$\setminus \setminus \setminus \setminus \setminus = \pi \sqrt{2} + 4 \pi \sqrt{2}$
$\setminus \setminus \setminus \setminus \setminus = 5 \sqrt{2} \pi$

Method 2

There is no need to use calculus for this problem, however if you are required to do so; then:

$V O R = \int \setminus 2 \pi y \setminus \mathrm{dS}$, where $\mathrm{dS} = \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \setminus \mathrm{dx}$

We need to take care at the point where our curve is below the $x$-axis as that are will be counted as negative (as with an a standard integral); Therefore,

On the interval; $0 \le x \le 1$

$y = 1 - x \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

$S A = 2 \pi \setminus {\int}_{0}^{1} \left(1 - x\right) \sqrt{1 + {\left(- 1\right)}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \setminus {\int}_{0}^{1} \left(1 - x\right) \sqrt{2} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus {\int}_{0}^{1} 1 - x \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus {\left[x - {x}^{2} / 2\right]}_{0}^{1}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus \left\{\left(1 - \frac{1}{2}\right) - \left(0 - 0\right)\right\}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \cdot \frac{1}{2}$
$\setminus \setminus \setminus \setminus \setminus = \pi \sqrt{2}$

On the interval; $1 \le x \le 3$

$y = x - 1 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$S A = 2 \pi \setminus {\int}_{1}^{3} \left(x - 1\right) \sqrt{1 + {\left(1\right)}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \setminus {\int}_{1}^{3} \left(x - 1\right) \sqrt{2} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus {\int}_{1}^{3} x - 1 \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus {\left[{x}^{2} / 2 - x\right]}_{1}^{3}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus \left\{\left(\frac{9}{2} - 3\right) - \left(\frac{1}{2} - 1\right)\right\}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \left(\frac{3}{2} + \frac{1}{2}\right)$
$\setminus \setminus \setminus \setminus \setminus = 4 \pi \sqrt{2}$

And when we add up these results, we get:

$S A = 5 \pi \sqrt{2}$, as before

WARNING NOTES

You have to be extremely careful, and not just blindly apply formula without thinking about the particular problem the formula is being applied to. If we did not account for the negativity of the function over the interval and just computed the formula for the entire interval we get:

On the interval; $0 \le x \le 3$

$y = 1 - x \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

$S A = 2 \pi \setminus {\int}_{0}^{3} \left(1 - x\right) \sqrt{1 + {\left(- 1\right)}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \setminus {\int}_{0}^{3} \left(1 - x\right) \sqrt{2} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus {\int}_{0}^{3} 1 - x \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus {\left[x - {x}^{2} / 2\right]}_{0}^{3}$
$\setminus \setminus \setminus \setminus \setminus = 2 \pi \sqrt{2} \setminus \left(- \frac{3}{2}\right)$
$\setminus \setminus \setminus \setminus \setminus = - 3 \pi \sqrt{2}$

Which is the net area with positive contribution, so here the smaller cone has positive contribution and the larger cone negative contribution towards the total surface area.