Area of Surface of Revolution

The area of the surface generated by revolving about the $x$-axis the arc $L$ of the curve $y=y(x)$ ($a\leq x\leq b$) is expressed by the integral
MATH
If a curve is represented parametrically or in polar coordinates, then it is sufficient to change the variable in the above formula, expressing appropriately the differential of the arc length.

Example 1

Find the area of the surface formed by revolving the astroid MATH about the $x$-axis.

Solution

Differentiating the equation of the astroid we get
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whence
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Then,
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Since the astroid is symmetric about the $y$-axis, in computing the area of the surface we may first assume $x\geq 0,$ and then double the result. In other words, the desired area $P$ is equal to
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Make the substitution
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MATH
Then
MATH

Example 2

Find the area of the surface generated by revolving about the x-axis a closed contour formed by the curves $y=x^{2}$ and $x=y^{2}.$

Solution. It is easy to check that the given parabolas intersect at the points $(0,0)$ and $(1,1).$ The sought-for area $P=P_{1}+P_{2}$, where the area $P_{1}$ is formed by revolving the arc $x=y^{2}$, and $P_{2}$ by revolving the arc $y=x^{2}$.

From the equation $y=x^{2}$ we get $y=\sqrt{x}$ and MATH.

Hence,
MATH

Now compute the area $P_{2}.$ We have MATH and
MATH

Thus
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Example 3

Find the area of the surface obtained by revolving a loop of the curve MATH about the $y$-axis.

Solution. The loop is described by a moving point as $y$ changes from $0$ to $3a.$ Differentiate with respect to $y$ both sides of the equation of the curve:


MATH
whence MATH Using the formula for computing the area of the surface of a solid of revolution about the $y$-axis, we have
MATH

Example 4

Compute the area of a surface generated by revolving about the $x$-axis an arc of the curve
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between the points of intersection of the curve and the $x$-axis.

Solution

Putting $y=0,\,$we find $t_{1}=0$, $t_{2}=\sqrt{3},$and $t_{3}=-\sqrt{3},$and hence MATH It follows that the curve intersects the $x$-axis at two points: $(0,0)$ and $(3,0).$ When $t$ is replaced by MATH This curve is thus symmetric about the x-axis.

To find the area of the surface it is sufficient to confine ourselves to the lower portion of the curve that corresponds to the variation of the parameter between $0$ and $+\sqrt{3}$. Differentiating with respect to $t,$ we find
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and the linear element
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Hence
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