Surface area of revolution around the x-axis and y-axis

 
 
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Formulas to find the surface area of revolution

We can use integrals to find the surface area of the three-dimensional figure that’s created when we take a function and rotate it around an axis and over a certain interval. 

The formulas we use to find surface area of revolution are different depending on the form of the original function and the axis of rotation.

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1. When the function is in the form y=f(x)y=f(x) and you’re rotating around the yy-axis, the interval is axba\leq{x}\leq{b} and the formula is

S=ab2πx1+(dydx)2 dxS=\int^b_a2\pi{x}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx

2. When the function is in the form y=f(x)y=f(x) and you’re rotating around the xx-axis, the interval is axba\leq{x}\leq{b} and the formula is

S=ab2πy1+(dydx)2 dxS=\int^b_a2\pi{y}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx

3. When the function is in the form x=g(y)x=g(y) and you’re rotating around the yy-axis, the interval is cydc\leq{y}\leq{d} and the formula is

S=cd2πx1+(dxdy)2 dyS=\int^d_c2\pi{x}\sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy

4. When the function is in the form x=g(y)x=g(y) and you’re rotating around the xx-axis, the interval is cydc\leq{y}\leq{d} and the formula is

S=cd2πy1+(dxdy)2 dyS=\int^d_c2\pi{y}\sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy

 
 

How to calculate surface area of revolution, whether you’re rotating around the x-axis or the y-axis


 
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Finding surface area of the rotation around the x-axis over an interval

Example

Find the area of the surface generated by rotating the function about the xx-axis over 0x30\leq{x}\leq3.

y=x3y=x^3

Since the equation is in the form y=f(x)y=f(x), and we’re rotating around the xx-axis, we’ll use the formula

S=ab2πy1+(dydx)2 dxS=\int^b_a2\pi{y}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx

We’ll calculate dy/dxdy/dx and then substitute it back into the equation.

dydx=3x2\frac{dy}{dx}=3x^2

S=032πx31+(3x2)2 dxS=\int^3_02\pi{x^3}\sqrt{1+\left(3x^2\right)^2}\ dx

S=032πx31+9x4 dxS=\int^3_02\pi{x^3}\sqrt{1+9x^4}\ dx

Using u-substitution and setting u=1+9x4u=1+9x^4 and du=36x3 dxdu=36x^3\ dx, we calculate

x=(u19)14x=\left(\frac{u-1}{9}\right)^{\frac14}

dx=136x3 dudx=\frac{1}{36x^3}\ du

dx=136[(u19)14]3 dudx=\frac{1}{36\left[\left(\frac{u-1}{9}\right)^{\frac{1}{4}}\right]^3}\ du

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The formulas we use to find surface area of revolution are different depending on the form of the original function and the axis of rotation.

Plugging these values back into the integral, we get

S=032π[(u19)14]3u136[(u19)14]3 duS=\int^3_02\pi{\left[\left(\frac{u-1}{9}\right)^{\frac{1}{4}}\right]^3}\sqrt{u}\frac{1}{36\left[\left(\frac{u-1}{9}\right)^{\frac{1}{4}}\right]^3}\ du

S=032π(u19)34u136(u19)34 duS=\int^3_02\pi{\left(\frac{u-1}{9}\right)^{\frac{3}{4}}}\sqrt{u}\frac{1}{36\left(\frac{u-1}{9}\right)^{\frac{3}{4}}}\ du

S=π1803(u19)34u1(u19)34 duS=\frac{\pi}{18}\int^3_0{\left(\frac{u-1}{9}\right)^{\frac{3}{4}}}\sqrt{u}\frac{1}{\left(\frac{u-1}{9}\right)^{\frac{3}{4}}}\ du

S=π1803u duS=\frac{\pi}{18}\int^3_0\sqrt{u}\ du

Integrate.

S=(π18)(23u32)03S=\left(\frac{\pi}{18}\right)\left(\frac{2}{3}u^{\frac{3}{2}}\right)\bigg|^3_0

S=π27u3203S=\frac{\pi}{27}u^{\frac{3}{2}}\bigg|^3_0

We’ll plug back in for uu, remembering that u=1+9x4u=1+9x^4, and then evaluate over the interval.

S=π27(1+9x4)3203S=\frac{\pi}{27}\left(1+9x^4\right)^{\frac{3}{2}}\bigg|^3_0

S=π27[1+9(3)4]32[π27(1+9(0)4)32]S=\frac{\pi}{27}\left[1+9(3)^4\right]^{\frac{3}{2}}-\left[\frac{\pi}{27}\left(1+9(0)^4\right)^{\frac{3}{2}}\right]

S=2,294.8S=2,294.8 square units

The surface area obtained by rotating y=x3y=x^3 around the xx-axis over the interval 0x30\leq{x}\leq3 is S=2,294.8S=2,294.8.

 
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