# How do you find the area of the surface generated by rotating the curve about the y-axis y=cx+d, a<=x<=b?

Jun 11, 2017

The surface area for $y = c x + d$ rotated around the $y$ axis for a given interval $x \in \left[a , b\right]$ is given by:

$S = \pi \left({b}^{2} - {a}^{2}\right) \sqrt{{c}^{2} + 1}$, $\text{ "" } c > 0$

(Note: If $c = 0$ and $a = 0$, then the surface area reduces to the area of a circle of radius $b$.)

If $x$ is in $\left[a , b\right]$ then assuming $c > 0$:

${y}_{\min} = c a + d$

${y}_{\max} = c b + d$

(If $c < 0$, then the revolution around the $y$ axis would produce the same surface area by symmetry of $y = | c | x + d$ compared to $y = - | c | x + d$. The value of $d$ does not matter, since the rotation is around the $y$ axis, and the shape will be the same regardless of the y-intercept.)

The surface area for a revolution around the $y$ axis is

$S = 2 \pi {\int}_{c a + d}^{c b + d} f \left(y\right) \sqrt{1 + {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2}} \mathrm{dy}$

Evaluating the derivative, we get:

${\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2} = {\left(\frac{d}{\mathrm{dy}} \left[\frac{y - d}{c}\right]\right)}^{2}$

$= \frac{1}{c} ^ 2$

Therefore:

$S = 2 \pi {\int}_{c a + d}^{c b + d} \frac{y - d}{c} \sqrt{1 + \frac{1}{c} ^ 2} \mathrm{dy}$

$= \frac{2 \pi \sqrt{1 + \frac{1}{c} ^ 2}}{c} {\int}_{c a + d}^{c b + d} \left(y - d\right) \mathrm{dy}$

$= \frac{2 \pi \sqrt{1 + \frac{1}{c} ^ 2}}{c} | \left[{y}^{2} / 2 - \mathrm{dc} \dot{y}\right] {|}_{c a + d}^{c b + d}$

$= \frac{2 \pi \sqrt{1 + \frac{1}{c} ^ 2}}{c} \left[{\left(c b + d\right)}^{2} / 2 - \mathrm{dc} \dot{c b + d}\right] - \left[{\left(c a + d\right)}^{2} / 2 - \mathrm{dc} \dot{c a + d}\right]$

$= \frac{2 \pi \sqrt{1 + \frac{1}{c} ^ 2}}{c} \left[{\left(c b + d\right)}^{2} / 2 - \mathrm{dc} \dot{c b + d} - {\left(c a + d\right)}^{2} / 2 + \mathrm{dc} \dot{c a + d}\right]$

$= \frac{2 \pi \sqrt{1 + \frac{1}{c} ^ 2}}{c} \left[\frac{{b}^{2} {c}^{2} + 2 b c d + {d}^{2}}{2} - b c d - {d}^{2} - \frac{{a}^{2} {c}^{2} + 2 a c d + {d}^{2}}{2} + a c d + {d}^{2}\right]$

$= \frac{2 \pi \sqrt{1 + \frac{1}{c} ^ 2}}{c} \left[\frac{{b}^{2} {c}^{2} + 2 b c d + \cancel{{d}^{2}} - {a}^{2} {c}^{2} - 2 a c d - \cancel{{d}^{2}}}{2} - b c d - \cancel{{d}^{2}} + a c d + \cancel{{d}^{2}}\right]$

$= \frac{2 \pi \sqrt{1 + \frac{1}{c} ^ 2}}{\cancel{c}} \left[\frac{\cancel{c} \left({b}^{2} c + 2 d \left(b - a\right) - {a}^{2} c\right)}{2} + \cancel{c} d \left(a - b\right)\right]$

$= 2 \pi \sqrt{1 + \frac{1}{c} ^ 2} \left[\frac{c \left({b}^{2} - {a}^{2}\right) + 2 d \left(b - a\right)}{2} + \frac{2 d \left(a - b\right)}{2}\right]$

$= 2 \pi \sqrt{1 + \frac{1}{c} ^ 2} \left[\frac{c \left({b}^{2} - {a}^{2}\right)}{2}\right]$

$= 2 \pi \sqrt{{c}^{2} + 1} \cdot \frac{{b}^{2} - {a}^{2}}{2}$

$= \boldsymbol{\textcolor{b l u e}{\pi \left({b}^{2} - {a}^{2}\right) \sqrt{{c}^{2} + 1}}}$

For example, let's rotate $y = 2 x + 3$ in the interval $\left[x = 0 , x = 2\right]$ about the $y$ axis using this formula and check our work.

S stackrel(?" ")(=) pi (2^2 - 0^2)sqrt(2^2 + 1)

stackrel(?" ")(=) 4sqrt5pi

In Wolfram Alpha, we get that it works!

Jun 12, 2017

$A = \pi \left({b}^{2} - {a}^{2}\right) \sqrt{1 + {c}^{2}}$

#### Explanation:

The function:

$y = c x + d$

represents a straight line. Rotating a straight line about $O y$ will form a cone, and so as we are rotating a closed interval $a \le x \le b$ we will generate a frustum. The surface area of a frustum is easily calculated using the formula:

$A = \pi \left(R + r\right) \sqrt{{\left(R - r\right)}^{2} + {h}^{2}}$

Where:

$R$ is the lower radius
$r$ is the upper radius
$h$ is the vertical height

When:

$x = a \implies y = c a + d$
$x = b \implies y = c b + d$

Thus we have:

$r = a , R = b$
$h = \left(c b + d\right) - \left(c a + d\right) = c b - c a$

Thus, the SA, is:

$A = \pi \left(a + b\right) \sqrt{{\left(b - a\right)}^{2} + {\left(c b - c a\right)}^{2}}$
$\setminus \setminus \setminus = \pi \left(a + b\right) \sqrt{{\left(b - a\right)}^{2} + {c}^{2} {\left(b - a\right)}^{2}}$
$\setminus \setminus \setminus = \pi \left(a + b\right) \sqrt{{\left(b - a\right)}^{2} \left(1 + {c}^{2}\right)}$
$\setminus \setminus \setminus = \pi \left(a + b\right) \left(b - a\right) \sqrt{1 + {c}^{2}}$
$\setminus \setminus \setminus = \pi \left({b}^{2} - {a}^{2}\right) \sqrt{1 + {c}^{2}}$