Volume of a Solid with Integrals

Question

What's the integral that will find the volume of the solid that is formed when the region bounded by the graphs of y = e^x, x = 1, and y = 1 is revolved around the line y = -2?

Answer 1

Mack, draw a diagram. The required area to be rotated lies below the curve y = e^x between x = 0 and x = 1.

This area is to be rotated around the horizontal line y = -2.

Cut a thin slice of thickness dx at the point X = x. This is a "washer", i.e., with a central hole (below the X-axis) of radius 2.

Therefore, its volume, dV = π ((y+2)² - 2²) dx

V = ∫[x = 0 to 1] (π (e^x+2)² - 2²) dx

= π ((1/3)(e^x+2)³ - 4x) |x = 0 to x = 1

= π ((e+2)³ - 4 - (e + 2) - 0)

= π ((e + 2)³ - e - 6)

= π (e³ + 6e² + 6e + 8 -e - 6)

= π (e³ + 6e² + 5e + 2) cubic units