volume of the solid generated when R is revolved about the x-axis.
Step 1: If there are two curves, f(x) and g(x), where f(x) is always above g(x), on the interval
[a, b], then the volume of the solid generated when the region is revolved about the x-axis is
found by using the method of washers.
We already know that f(x) is above g(x) on the interval, so the integral we need to evaluate is
(c) Set up, but do not integrate, an integral expression, in terms of a single variable, for the
volume of the solid generated when R is revolved about the line x = −1.
Step 1: Now we have to revolve the area around a vertical axis. If there are two curves, f(x)
and g(x), where f(x) is always above g(x), on the interval [a, b], then the volume of the solid
generated when the region is revolved about the y-axis is found by using the method of shells.
2π x[f(x) − g(x)] dx
When we are rotating around a vertical axis, we use the same formula as when we rotate
around the y-axis, but we have to account for the shift away from x = 0. Here we have a curve
that is 1 unit farther away from the line x = − 1 than it is from the y-axis, so we add 1 to the
radius of the shell (for a more detailed explanation of shifting axes, see the unit on finding the
volume of a solid of revolution). This gives us the equation,
- Consider the equation x^2 − 2xy + 4y^2 = 52.
(a) Write an expression for the slope of the curve at any point (x, y).
Step 1: The slope of the curve is just the derivative. But here we have to use implicit
differentiation to find the derivative. If we take the derivative of each term with respect to x,
we get
2 x − + 8 y = 0
