Surface area of revolution of a parametric curve, horizontal axis
Formula for the surface area of revolution
The surface area of the solid created by revolving a parametric curve around the ???x???-axis is given by
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???S_x=\int^b_a 2\pi{y}\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 ???f(t)???
where ???g'(t)??? is the derivative of the curve ???g(t)???
How to find the surface area of revolution of a parametric curve
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Finding surface area of revolution of a parametric curve when we rotate around the x-axis
Example
Find the surface area of revolution of the solid created when the parametric curve is rotated around the ???x???-axis over the interval ???0\le t\le\frac{\pi}{2}???.
???x=\cos^3{t}???
???y=\sin^3{t}???
We’ll call the parametric equations
???f(t)=\cos^3{t}???
???g(t)=\sin^3{t}???
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)=-3\cos^2{t}\sin{t}???
???g'(t)=3\sin^2{t}\cos{t}???
Now we’ll plug into the formula for the surface area of revolution.
???S_x=\int^{\frac{\pi}{2}}_0 2\pi{\left(\sin^3{t}\right)}\sqrt{\left(-3\cos^2{t}\sin{t}\right)^2+\left(3\sin^2{t}\cos{t}\right)^2}\ dt???
???S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\sqrt{9\cos^4{t}\sin^2{t}+9\sin^4{t}\cos^2{t}}\ dt???
???S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\sqrt{9\sin^2{t}\cos^2{t}\left(\cos^2{t}+\sin^2{t}\right)}\ dt???
We need to find both derivatives before we can plug into the formula.
Since ???\sin^2{t}+\cos^2{t}=1???, we get
???S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\sqrt{9\sin^2{t}\cos^2{t}\left(1\right)}\ dt???
???S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\sqrt{9\sin^2{t}\cos^2{t}}\ dt???
???S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\left(3\sin{t}\cos{t}\right)\ dt???
???S_x=6\pi\int^{\frac{\pi}{2}}_0\sin^4{t}\cos{t}\ dt???
We’ll use u-substitution, letting
???u=\sin{t}???
???du=\cos{t}\ dt???
We’ll make the substitution.
???S_x=6\pi\int^{x=\frac{\pi}{2}}_{x=0}u^4\ du???
???S_x=\frac{6\pi}{5}u^5\Big|^{x=\frac{\pi}{2}}_{x=0}???
Back-substituting for ???u???, we get
???S_x=\frac{6\pi}{5}\sin^5{t}\Big|^{\frac{\pi}{2}}_0???
???S_x=\left(\frac{6\pi}{5}\sin^5{\frac{\pi}{2}}\right)-\left(\frac{6\pi}{5}\sin^5{0}\right)???
???S_x=\frac{6\pi}{5}(1)^5-\frac{6\pi}{5}(0)^5???
???S_x=\frac{6\pi}{5}???
