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17Calculus Parametric Equations - Volume

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Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
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On this page we discuss how to calculate a volume described by parametric equations.

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

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

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.

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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

video by Krista King Math

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Practice Instructions

Unless otherwise instructed, calculate the volume of revolution when the given curve is revolved about the given axis. Give your answer in exact form.

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