17Calculus Parametric Equations - Volume
17Calculus
On this page we discuss how to calculate a volume described by parametric equations.
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Equations
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 \( t=t_1 \).
Revolution About the x-axis |
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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 \( dx = [dx/dt] dt \) |
Revolution About the y-axis |
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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
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.
Don't forget the \(\pi\) out in front of the integrals.
Practice
Unless otherwise instructed, calculate the volume of revolution when the given curve is revolved about the given axis. Give your answer in exact form.
\(x = t^3\), \(y = 2t^2+1\), \(-1 \leq t \leq 1\); \(x\)-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
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