- Let R be the region enclosed by the graphs of y = , y = x^2 , and the lines x = 0 and x = 1.
(a) Find the area of R.
First, determine which curve is more positive (f(x)), and set up your integral for area between
curves: A = (f(x) − g(x)) dx. For this problem, f(x) = , so the integral is A =
dx = 2ln|2| − ≈ 1.05296.
(b) Find the volume of the solid generated when R is revolved about the x-axis.
You can use the washer method to find the volume: V = π [(f(x))^2 − (g(x))^2 ] dx. Thus, V =
dx = 1.8π.
(c) Set up, but do not evaluate, the expression for the volume of the solid generated when R is
revolved around the line x = 2.
Here, use the cylindrical shells method: V = 2π x(f(x) − g(x)) dx. Adjust the axis of rotation
since we are revolving around the line x = 2. Because x = 2 is more positive than the x-axis,
we set up the integral with (2 − x), not x. Thus, the integral is V = 2π
dx.
- Consider the equation x^3 + 2x^2 y + 4y^2 = 12.
(a) Write an equation for the slope of the curve at any point (x, y).
Use implicit differentiation to find the first derivative, which is the slope of the curve at any
point (x, y):
x^3 + 2x^2 y + 4y^2 = 12
3 x^2 + 2 + 8y = 0
3 x^2 + 2x^2 + 4xy + 8y = 0
