Mathematics | Area of the surface of solid of revolution

Difficulty Level : Medium
Last Updated : 10 May, 2020

Consider a plane y=f(x) in the x-y plane between ordinates x=a and x=b. If a certain portion of this curve is revolved about an axis, a solid of revolution is generated.

We can calculate the area of this revolution in various ways such as:

Cartesian Form:
- Area of solid formed by revolving the arc of curve about x-axis is-
  $S= \int_{x=a}^{x=b} 2\pi y\sqrt{1+(\frac{dy}{dx})^2}dx$
- Area of revolution by revolving the curve about y axis is-
  $S= \int_{y=c}^{y=d} 2\pi x \sqrt{1+(\frac{dx}{dy})^2}dy$

Parametric Form: $x=x(t), y=y(t)$
- About x-axis:
  $S=\int_{t=t_{1}}^{t=t_{2}} 2\pi y(t) \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$
- About y-axis:
  $S=\int_{t=t_{1}}^{t=t_{2}} 2\pi x(t) \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Polar Form: r=f(θ)
- About the x-axis: initial line $\theta = \frac{\pi}{2}$
  $S= \int_{\theta=\theta_1}^{\theta _2}2\pi y\frac{ds}{d\theta}d\theta$
  $=\int_{\theta=\theta_1}^{\theta _2}2\pi (r sin\theta) \sqrt{r^2+(\frac {dr}{d\theta})^2}d\theta$
  Here replace r by f(θ)
- About the y-axis:
  $S= \int_{\theta=\theta_1}^{\theta _2}2\pi x\frac{ds}{d\theta}d\theta$
  $=\int_{\theta=\theta_1}^{\theta _2}2\pi (r cos\theta) \sqrt{r^2+(\frac {dr}{d\theta})^2}d\theta$
  Here replace r by f(θ)

About any axis or line L: $S= \int 2\pi (PM) ds$ where PM is the perpendicular distance of a point P of the curve to the given axis.
- Limits for x: x = a to x = b
  $S=\int_{x=a}^{x=b} 2\pi (PM)\sqrt{1+(\frac{dy}{dx})^2}dx$
  Here PM is in terms of x.
- Limits for y: y = c to y = d
  $S= \int_{y=c}^{y=d} 2\pi (PM)\sqrt{1+(\frac{dx}{dy})^2}dy$
  Here PM is in terms of y.

Example:
Find the area of the solid of revolution generated by revolving the parabola $y^2=4ax, 0\leq x \leq 3a$ about the x-axis.
Explanation:
Now we are given with the Cartesian form of the equation of parabola and the parabola has been rotated about the x-axis. Hence we use the formula for revolving Cartesian form about x-axis which is:

$S= \int_{x=a}^{x=b} 2\pi y\sqrt{1+(\frac{dy}{dx})^2}dx$

Here $y^2= 4ax$ . Now we need to calculate dy/dx

Differentiating w.r.t x we get:

$2yy'= 4a$

$y'=\frac{2a}{y}$

$1+(y')^2=1+\frac {4a^2}{y^2}=\frac{y^2+4a^2}{y^2}$

Using $y^2=4ax$

$\sqrt {1+(y')^2}=\sqrt{\frac{4ax+4a^2}{y^2}}=\frac{2\sqrt a}{y}\sqrt {a+x}$

Now we are provided with limits of x as x=0 to x=3. Plugging our calculated values in the above formula we get:

$S=\int_{0}^{3a} 2\pi y.{\frac{2\sqrt a}{y}\sqrt {a+x}}dx$

$=2\pi\int_{0}^{3a} y.{\frac{2\sqrt a}{y}\sqrt {a+x}}dx$

$=4\pi\sqrt a\int_{0}^{3a}\sqrt {a+x}$

$=4\pi\sqrt a\int_{0}^{3a}\frac{2}{3}(x+a)^{3/2}\Biggr|_{0}^{3a}$

$=\frac{8}{3}\pi\sqrt a ((4a)^{3/2}-(a)^{3/2})$

$=\frac{8}{3}\pi\sqrt a.a^{3/2}(8-1)$

$=\frac{56\pi a^2}{3} sq. units$

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