
William W. answered 07/19/21
Math and science made easy - learn from a retired engineer
Sketch:
At any random spot between x = 0 and x = 2 the cross section of the solid is a circle. The radius of that circle is the function value of y = e-x so the area of that circle is A = π(rπ2) = π(e-x)2 = πe-2x so the volume is:
V(x) = 0∫2 πe-2x dx = π•(-1/2)e-2x evaluated between 0 and 2 = (-π/2)(e-4) - -π/2(1) = π/2 - π/(2e4) ≈ 1.542 cubic units
For the second problem, you have a circle rotated around a center line giving a doughnut shape. Integrating in the y-direction provides a "washer" for each cross section. Calculate the inside radius as a function of "y" and then the outside radius as a function of "y", then use the fact that the area of a "washer" is π(ro2 - ri2). Hint: The inside radius at y = 0 is b - a and the inside radius at y = a is b while the outside radius at y = 0 is a + b and the outside radius at y = a is b. The Volume is 2•0∫a Awasher dy