## 17Calculus Parametric Equations - Volume

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

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 Wikipedia - Parametric Derivative Wikipedia - Arc Length Pauls Online Notes - Arc Length with Parametric Equations Wikipedia - Solid of Revolution

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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