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How to find the volume of revolution of a parametric curve

Formulas for the volume of revolution of a parametric curve

In the same way that we could find the volume of a three-dimensional object generated by rotating a two-dimensional area around an axis when we studied applications of integrals, we can find the volume of revolution generated by revolving the area enclosed by two parametric curves.

The formulas we use to find the volume of revolution for a parametric curve are:

Rotation around the ???x???-axis:

???V_x=\int^\beta_\alpha\pi{y^2}\frac{dx}{dt}\ dt???

Rotation around the ???y???-axis":

???V_y=\int^\beta_\alpha\pi{x^2}\frac{dy}{dt}\ dt???

How to calculate the volume of revolution for a parametric curve


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Find the volume of revolution of the parametric curve, if we’re rotating around the y-axis

Example

Find the volume of revolution of the parametric curve for ???0\le{t}\le2???, rotated around the ???y???-axis.

???x=3t^2+4???

???y=t^4???

Since we’re rotating around the ???y???-axis, we’ll use the formula

???V_y=\int^\beta_\alpha\pi{x^2}\frac{dy}{dt}\ dt???

The problem gave the interval ???0\le{t}\le2???, so ???\alpha=0??? and ???\beta=2???. Now we need to find ???dy/dt??? so that we can plug it into the volume formula.

???y=t^4???

???\frac{dy}{dt}=4t^3???

Plugging everything into the volume formula, we get

???V_y=\int^2_0\pi\left(3t^2+4\right)^2\left(4t^3\right)\ dt???

???V_y=4\pi\int^2_0t^3\left(3t^2+4\right)^2\ dt???

???V_y=4\pi\int^2_0t^3\left(9t^4+24t^2+16\right)\ dt???

???V_y=4\pi\int^2_09t^7+24t^5+16t^3\ dt???

???V_y=4\pi\left(\frac{9t^8}{8}+\frac{24t^6}{6}+\frac{16t^4}{4}\right)\bigg|^2_0???

???V_y=4\pi\left(\frac{9t^8}{8}+4t^6+4t^4\right)\bigg|^2_0???

???V_y=4\pi\left[\frac{9(2)^8}{8}+4(2)^6+4(2)^4\right]-4\pi\left[\frac{9(0)^8}{8}+4(0)^6+4(0)^4\right]???

???V_y=4\pi\left[9(2)^5+256+64\right]???

???V_y=2,432\pi???


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