The area of the surface generated by revolving about the
-axis
the arc
of the curve
()
is expressed by the integral

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.

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

**Solution**

Differentiating the equation of the astroid we get

whence

Then,

Since the astroid is symmetric about the
-axis,
in computing the area of the surface we may first assume
and then double the result. In other words, the desired area
is equal to

Make the substitution

Then

Find the area of the surface generated by revolving about the x-axis a closed contour formed by the curves and

Solution. It is easy to check that the given parabolas intersect at the points and The sought-for area , where the area is formed by revolving the arc , and by revolving the arc .

From the equation we get and .

Hence,

Now compute the area
We have
and

Thus

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

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

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

Compute the area of a surface generated by revolving about the
-axis
an arc of the curve

between the points of intersection of the curve and the
-axis.

**Solution**

Putting we find , and and hence It follows that the curve intersects the -axis at two points: and When is replaced by 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
and
.
Differentiating with respect to
we find

and the linear element

Hence