Question

Consider the region satisfying the inequalities.y ≤ e−x, y ≥ 0, x ≥ 0a) Find area of regionb) Find the volume of the solid generated by revolving the region about the x-axis.c) Find the volume of the solid generated by revolving the region about the y-axis.

Answer 1

The area of the region is 1 square unit. The volume of the solid generated by revolving the region about the x-axis can be found by integrating π(y^2) dx from x = 0 to x = ∞.

Step-by-step explanation:

To find the area of the region, we need to find the intersection points between the two curves. In this case, the curves are y = e^(-x) and y = 0. Since y ≥ 0, the region will lie between the x-axis and the curve y = e^(-x). The intersection point is where y = 0, which occurs at x = 0. To find the area, we integrate y = e^(-x) from x = 0 to x = ∞:

A = ∫0∞ e^(-x) dx = [-e^(-x)]0∞ = -[e^0 - 0]
= -[1 - 0] = 1

The area of the region is 1 square unit.

To find the volume of the solid generated by revolving the region about the x-axis, we use the disk method. The radius of each disk is given by y = e^(-x), and the height of each disk is given by dx. The volume can be found by integrating π(y^2) dx from x = 0 to x = ∞:

V = π∫0∞ (e^(-x))^2 dx = π∫0∞ e^(-2x) dx

Answer 2

• Revolving about the
  `x`-axis:

Using the disk method, the volume is

`\displaystyle\pi\int_0^\infty e^(-2x)\,\mathrm dx=\boxed{\frac\pi2}`

Alternatively, using the shell method, the volume is

`\displaystyle2\pi\int_0^1y(-\ln y)\,\mathrm dy=\frac\pi2`

• Revolving about the
  `y`-axis:

Using the shell method, the volume is

`\displaystyle2\pi\int_0^\infty xe^(-x)\,\mathrm dx=\boxed{2\pi}`

Alternatively, using the disk method, the volume is

`\displaystyle\pi\int_0^1(-\ln x)^2\,\mathrm dx=2\pi`