
Juan M. answered 04/30/23
Professional Math and Physics Tutor
31. y = e^(-x^2), y = 0, x = -1, x = 1:
(a) About the x-axis:
To find the volume of the solid generated by rotating the region bounded by the given curves about the x-axis, we can use the disk method. The volume V can be found using the disk method:
V = pi * ∫(from -1 to 1) (e^(-x^2))^2 dx
V = pi * ∫(from -1 to 1) e^(-2x^2) dx
Use a calculator to evaluate the integral:
V ≈ pi * 0.88208 = 2.76894 * pi
(b) About y = -1:
To find the volume of the solid generated by rotating the region bounded by the given curves about the line y = -1, we can use the washer method. The volume V can be found using the washer method:
V = pi * ∫(from -1 to 1) [(1 + e^(-x^2))^2 - 1^2] dx
V = pi * ∫(from -1 to 1) (1 + 2e^(-x^2) + e^(-2x^2) - 1) dx
V = pi * ∫(from -1 to 1) (2e^(-x^2) + e^(-2x^2)) dx
Use a calculator to evaluate the integral:
V ≈ pi * 1.76416 = 5.53787 * pi
33. x^2 + 4y^2 = 4:
(a) About y = 2:
First, we need to solve for x in terms of y:
x^2 = 4 - 4y^2
x = ±sqrt(4 - 4y^2)
Since the region of interest is symmetric about the y-axis, we can consider only the positive square root and multiply the result by 2:
x = sqrt(4 - 4y^2)
Now, we need to find the limits of integration. The ellipse x^2 + 4y^2 = 4 intersects y = 2 when:
x^2 + 4(2)^2 = 4
x^2 = -12 (no solution)
Thus, the limits of integration are y = -1/2 to y = 1/2. The volume V can be found using the washer method:
V = 2 * pi * ∫(from -1/2 to 1/2) (sqrt(4 - 4y^2))^2 dy
V = 2 * pi * ∫(from -1/2 to 1/2) (4 - 4y^2) dy
Use a calculator to evaluate the integral:
V ≈ 2 * pi * 2 = 4 * pi
(b) x = 2:
The region of interest is symmetric about the x-axis, so the volume is 0.
53. A tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5 cm:
The volume of a tetrahedron with mutually perpendicular edges can be found using the formula:
V = (1/3) * base_area * height
In this case, the edges are the base and the height, so the formula becomes:
V = (1/3) * 3 * 4 * 5
V = 20 cm³