On this page we discuss integration and volume using parametric equations.
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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 xaxis 

When revolved about the xaxis, 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 yaxis 

Similar to the xaxis 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 xaxis, the integrand contains \(Y(t)\) squared. And we have a similar situation for revolution about the yaxis. This may seem somewhat counterintuitive 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 xaxis.
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 xaxis.
Solution
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