## 17Calculus Parametric Equations - Volume

##### 17Calculus

If we have a parametric curve defined as $$x = X(t)$$ and $$y = Y(t)$$, we can determine the volume of the solid object defined by revolving this curve about an axis. We will limit our curve from $$t=t_0$$ to $$t=t_1$$.

When revolved about the x-axis, the integral we use to calculate volume is $V_x = \pi \int_{t_0}^{t_1}{[Y(t)]^2 [dx/dt] dt}$ This is sometimes written $V_x = \pi \int_{t_0}^{t_1}{y^2 ~ dx}$ where $$dx = [dx/dt] dt$$

Similar to the x-axis integral, we use this integral to calculate volume. $V_y = \pi \int_{t_0}^{t_1}{[X(t)]^2 [dy/dt] dt}$ also written as $V_y = \pi \int_{t_0}^{t_1}{x^2 ~ dy}$ where $$dy = [dy/dt] dt$$

Notes
1. Notice that when revolving about x-axis, the integrand contains $$Y(t)$$ squared. And we have a similar situation for revolution about the y-axis. This may seem somewhat counter-intuitive but it makes sense when you know where the equations come from.
2. Don't forget the $$\pi$$ out in front of the integrals.

Practice

Calculate the volume of revolution when the curve $$x = t^3$$, $$y = 2t^2+1$$, $$-1 \leq t \leq 1$$ is rotated about the x-axis.

Problem Statement

Calculate the volume of revolution when the curve $$x = t^3$$, $$y = 2t^2+1$$, $$-1 \leq t \leq 1$$ is rotated about the x-axis.

Solution

### Krista King Math - 466 video solution

video by Krista King Math

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