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Wize University Calculus 2 Textbook > Applications of Integrals

Surface Area of Revolution

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Surface Area of Revolution

We can find the surface area of an object by revolving the function about an axis and multiplying it by our arc length factor.


Surface Area

If the curve y=f(x)y=f(x) is revolved about the x-axis on the interval [a,b][a,b], then the surface area of the resulting solid is found as
SA=∫ab2πf(x)1+[f′(x)]2 dx\boxed{SA = \displaystyle\int_a^b 2\pi f(x) \sqrt{1+\left[f'(x)\right]^2} \ dx}
If the curve is revolved about the y-axis we have

SA=∫ab2πg(y)1+[g′(y)]2dy \boxed{SA=\int_a^b2\pi g(y)\sqrt{1+[g'\left(y\right)]^2}dy}


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Example: Surface Area of Revolution

Find the surface area of the solid formed by revolving f(x)=xf(x)=\sqrt{x} about the x-axis on [0,1][0,1].

Noting that f′(x)=12xf'(x)=\dfrac{1}{2\sqrt{x}}, we have
SA=∫012πx1+(12x)2 dx=∫012πx1+14x dx=∫012πx4x+14x dx=∫012πx⋅4x+14x dx=∫012π4x+14 dx=∫01π4x+1 dx\begin{array}{rl} SA&=\displaystyle\int_0^1 2\pi \sqrt{x}\sqrt{1+\left(\dfrac{1}{2\sqrt{x}}\right)^2}\ dx \\[+2em] &=\displaystyle\int_0^1 2\pi \sqrt{x}\sqrt{1+\dfrac{1}{4x}}\ dx \\[+2em] &= \displaystyle\int_0^1 2\pi \sqrt{x}\sqrt{\dfrac{4x+1}{4x}} \ dx \\[+2em] &=\displaystyle\int_0^1 2\pi \sqrt{x\cdot \dfrac{4x+1}{4x}}\ dx \\[+2em] &=\displaystyle\int_0^1 2\pi \sqrt{\dfrac{4x+1}{4}}\ dx \\[+2em] &=\displaystyle\int_0^1 \pi\sqrt{4x+1}\ dx \end{array}
Letting u=4x+1u=4x+1, we have du=4 dx→14du=dx, u(0)=1,u(1)=5du=4\ dx \rightarrow \dfrac{1}{4}du = dx, \ u(0)=1, u(1)=5:

SA=π4∫15u du=π4⋅u3/23/2∣15=π2⋅u3/23∣15=π6(53/2−1)\begin{array}{rl} SA & = \dfrac{\pi}{4} \displaystyle\int_1^5 \sqrt{u}\ du \\[+2em] &=\dfrac{\pi}{4}\cdot \dfrac{u^{3/2}}{3/2}\bigg|_1^5 \\[+2em] &=\dfrac{\pi}{2}\cdot \dfrac{u^{3/2}}{3}\big|_1^5 \\[+2em] &=\dfrac{\pi}{6}(5^{3/2}-1) \end{array}

Practice: Surface Area of Revolution

Find the surface area of a right circular cone of height 4, radius 2.