Final answer:
To find the volume of the solid generated by revolving the region bounded by the graphs of the equations y = e^−x, y = 0, x = 0, x = 9 about the x-axis using the disk method, we need to integrate the area of a cross-section of the region and then rotate it around the x-axis. The volume of the solid generated is π / 3.
Step-by-step explanation:
To find the volume of the solid generated by revolving the region bounded by the graphs of the equations y = e^−x, y = 0, x = 0, x = 9 about the x-axis using the disk method, we need to integrate the area of a cross-section of the region and then rotate it around the x-axis. Since the boundaries of the region are given in terms of x, we need to express the equations in terms of y.
The equation y = e^−x can be rewritten as x = -ln(y). Now, we can find the bounds of integration by solving for the values of y where the region intersects the x-axis. Setting y = 0, we get x = -ln(0) which is undefined, but since x = 0 and x = 9 are given as boundaries, we can consider the interval [0, 9].
The volume of the solid can be found using the formula for the volume of a disk:
V = π * ∫(R(y))^2 dy
where R(y) represents the radius of each disk. In this case, the radius R(y) is given by the equation R(y) = -ln(y). Plugging this into the formula:
V = π * ∫((-ln(y))^2) dy
To evaluate the integral, we need to find the antiderivative and apply the limits of integration:
V = π * [((y^3)/(3*(ln(y))^2)) + (2*(y^2)/(ln(y))) - (2*y/ln(y))] evaluated from y = 0 to y = e^0 = 1.
After substituting the limits, we get:
V = π * [(1/3) + 2 - 2] = π * (1/3)
So, the volume of the solid generated is π / 3.