Finding surface area of revolution of a parametric curve around a vertical axis

The formula for surface area of revolution of a parametric curve

The surface area of the solid created by revolving a parametric curve around the $y$-axis is given by

$S_x=\int^b_a 2\pi{x}\sqrt{\left[f'(t)\right]^2+\left[g'(t)\right]^2}\ dt$

where the curve is defined over the interval $[a,b]$,

where $f'(t)$ is the derivative of the curve $x=f(t)$

where $g'(t)$ is the derivative of the curve $y=g(t)$

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How to find surface area of revolution with a parametric curve around a vertical axis of revolution

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Finding surface area of the parametric curve rotated around the y-axis

Example

Find the surface area of revolution of the solid created when the parametric curve is rotated around the given axis over the given interval.

$x=2t^2$

$y=2t^3$

for $0\le{t}\le3$, rotated around the $y$-axis

We’ll call the parametric equations

$f(t)=2t^2$

$g(t)=2t^3$

The limits of integration are defined in the problem, but we need to find both derivatives before we can plug into the formula.

$f'(t)=4t$

$g'(t)=6t^2$

Now we’ll plug into the formula for the surface area of revolution.

$S_y=\int^3_02\pi\left(2t^2\right)\sqrt{\left(4t\right)^2+\left(6t^2\right)^2}\ dt$

$S_y=\int^3_04\pi{t^2}\sqrt{16t^2+36t^4}\ dt$

$S_y=\int^3_04\pi{t^2}\sqrt{4t^2\left(4+9t^2\right)}\ dt$

$S_y=\int^3_0 8\pi{t^3}\sqrt{4+9t^2}\ dt$

$S_y=8\pi\int^3_0t^3\sqrt{4+9t^2}\ dt$

We use different formulas to find the surface area of revolution of a parametric curve, depending on whether the axis of revolution is horizontal or vertical.

We’ll use u-substitution, letting

$u=4+9t^2$

$9t^2=u-4$

$t^2=\frac{u-4}{9}$

$du=18t\ dt$

$dt=\frac{du}{18t}$

We’ll make the substitution.

$S_y=8\pi\int^{t=3}_{t=0}t^3\sqrt{u}\ \frac{du}{18t}$

$S_y=\frac{8\pi}{18}\int^{t=3}_{t=0}t^2\sqrt{u}\ du$

$S_y=\frac{8\pi}{18}\int^{t=3}_{t=0}\frac{u-4}{9}\sqrt{u}\ du$

$S_y=\frac{8\pi}{18}\int^{t=3}_{t=0}\left(\frac{u}{9}-\frac49\right)u^{\frac12}\ du$

$S_y=\frac{8\pi}{18}\int^{t=3}_{t=0}\frac{u^{\frac32}}{9}-\frac{4u^{\frac12}}{9}\ du$

$S_y=\frac{8\pi}{162}\int^{t=3}_{t=0}u^{\frac32}-4u^{\frac12}\ du$

$S_y=\frac{4\pi}{81}\left(\frac25u^{\frac52}-\frac83u^{\frac32}\right)\bigg|^{t=3}_{t=0}$

Back-substituting for $u$, we get

$S_y=\frac{4\pi}{81}\left[\frac25\left(4+9t^2\right)^{\frac52}-\frac83\left(4+9t^2\right)^{\frac32}\right]\bigg|^3_0$

Evaluate over the interval.

$S_y=\frac{4\pi}{81}\left[\frac25\left(4+9(3)^2\right)^{\frac52}-\frac83\left(4+9(3)^2\right)^{\frac32}\right]-\frac{4\pi}{81}\left[\frac25\left(4+9(0)^2\right)^{\frac52}-\frac83\left(4+9(0)^2\right)^{\frac32}\right]$

$S_y=\frac{4\pi}{81}\left[\frac25\left(4+81\right)^{\frac52}-\frac83\left(4+81\right)^{\frac32}\right]-\frac{4\pi}{81}\left[\frac25\left(4+0\right)^{\frac52}-\frac83\left(4+0\right)^{\frac32}\right]$

$S_y=\frac{4\pi}{81}\left[\frac25\left(85\right)^{\frac52}-\frac83\left(85\right)^{\frac32}-\frac25\left(4\right)^{\frac52}+\frac83\left(4\right)^{\frac32}\right]$

$S_y=\frac{4\pi}{81}\left[\frac25\left[\left(85\right)^5\right]^\frac12-\frac83\left[\left(85\right)^3\right]^\frac12-\frac25\left[\left(4\right)^{\frac12}\right]^5+\frac83\left[\left(4\right)^{\frac12}\right]^3\right]$

$S_y=\frac{4\pi}{81}\left[\frac25\left[85\left(85\right)^4\right]^\frac12-\frac83\left[85\left(85\right)^2\right]^\frac12-\frac25(2)^5+\frac83(2)^3\right]$

$S_y=\frac{4\pi}{81}\left[\frac25\left[\left(85\right)^2\sqrt{85}\right]-\frac83\left[85\sqrt{85}\right]-\frac25(32)+\frac83(8)\right]$

$S_y=\frac{4\pi}{81}\left[\frac{2\left(85\right)^2\sqrt{85}}{5}-\frac{680\sqrt{85}}{3}-\frac{64}{5}+\frac{64}{3}\right]$

$S_y=\frac{4\pi}{81}\left[\frac{2\cdot5\cdot5\cdot17\cdot17\cdot\sqrt{85}}{5}-\frac{680\sqrt{85}}{3}-\frac{64}{5}+\frac{64}{3}\right]$

$S_y=\frac{4\pi}{81}\left[2,890\sqrt{85}-\frac{680\sqrt{85}}{3}-\frac{64}{5}+\frac{64}{3}\right]$

$S_y=\frac{4\pi}{81}\left[2,890\sqrt{85}-\frac{64}{5}+\frac{64-680\sqrt{85}}{3}\right]$

Find a common denominator.

$S_y=\frac{4\pi}{81}\left[\frac{43,350\sqrt{85}}{15}-\frac{192}{15}+\frac{320-3,400\sqrt{85}}{15}\right]$

$S_y=\frac{4\pi}{81}\left(\frac{43,350\sqrt{85}-3,400\sqrt{85}-192+320}{15}\right)$

$S_y=\frac{4\pi}{81}\left(\frac{39,950\sqrt{85}+128}{15}\right)$

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