Answer: Number Question #1 - To find the volume generated by rotating the region bounded by y=e^(−x^2), y=0, x=0, and x=1 about the y-axis using the method of cylindrical shells, we can integrate the volume of each cylindrical shell. The radius of each shell will be x, and the height will be e^(−x^2).
Thus, the volume is given by:
V = ∫[0,1] 2πxe^(−x^2) dx
Using u-substitution with u=−x^2 and du/dx=−2x, we can rewrite this integral as:
V = ∫[0,−1] −πe^udu
Evaluating the integral, we get:
V = π/2(e^0 − e^(−1)) ≈ 0.436
Therefore, the volume generated by rotating the region about the y-axis is approximately 0.436 cubic units.
Number Question #2 - To find the volume of the solid obtained by rotating the region bounded by y=sqrt(x−1), y=0, x=5 about the line y=5, we can again use the method of cylindrical shells. In this case, the radius of each shell will be 5−y, and the height will be x−1.
Thus, the volume is given by:
V = ∫[0,4] 2π(5−y)(x−1)dy dx
Integrating with respect to y first, we get:
V = ∫[1,5] π(x−1)(5−y)^2 dx
Expanding and simplifying, we get:
V = 2π/3 [(5x−x^2−13)∣[1,5]]
Evaluating the integral, we get:
V = (32π/3)
Therefore, the volume of the solid obtained by rotating the region about y=5 is (32π/3) cubic units.
Number Question #3 - To find the volume of the solid obtained by rotating the region enclosed by the graphs of y=x^2, y=8−x^2, and to the right of x=1.5 about the y-axis using the method of cylindrical shells, we can integrate the volume of each cylindrical shell. The radius of each shell will be x, and the height will be (8−x^2)−x^2=8−2x^2.
Thus, the volume is given by:
V = ∫[1.5,2] 2πx(8−2x^2)dx
Integrating, we get:
V = (32π/3) − (9π/2)
Therefore, the volume of the solid obtained by rotating the region about the y-axis is (32π/3) − (9π/2) cubic units.