1. Volume = 1.

1. The area is A(x) = base ⋅ height = x⋅e^x. The volume is V = ∫₀¹A(x) dx = ∫₀¹ x⋅e^x dx= (x⋅e^x – e^x)|₀¹ = 1 where x⋅e^x was integrated by parts, using u= x and dv= e^xdx.

2. Find the volume of the solid obtained by rotating the curve y= x², -1 ≤ x ≤ 1, about the line y= -2.

Then A(x) = π⋅(2 + x2)2 = π⋅(4 + 4×2 + x4), and so the volume is V = ∫-11π⋅(4 + 4×2 + x4) dx = π⋅(4x + (4/3)x3 + (1/5)x5)|-11 = π⋅166/15.

1. Volume = π⋅7/15.

1. Rotating the curve y = 2x – 2x² about the line y = 1 gives cross-sections that are circles of radius 1 – (2x – 2x²)

Then A(x) = π⋅(1 – (2x – 2x²)²= π⋅(1 – 4x + 8x² – 8x³ + 4x⁴), and so the volume is V = ∫₀¹π⋅(1 – 4x + 8x² – 8x³ + 4x⁴) dx = π⋅(x – 4(x²/2) + 8(x³/3) – 8(x⁴/4) + 4(x⁵/5)|₀¹ = π⋅7/15.

1. The answers are the following: (a) Volume = 4π/5. (b) Volume = 6π/7. (c) Volume = 2nπ/(2n+1).

1. (a) Rotating the region between y = 1 and y = x² around the x-axis gives an annular cross-section, the region between a circle of radius 1 and a circle of radius x².

Then A(x) = π⋅(1² – (x²)(²

– (x²)(²) = π⋅(1 – x⁴), and so the volume is V = ∫₀¹π⋅(1 – x⁴) dx = π⋅(1 – x⁵/5)|₀¹= 4π/5. (b) Here the radius of the inner circle is x3, so A(x) = π⋅(1 – (x³)2) and V = ∫₀¹π⋅(1 – x⁶) dx = π⋅(1 – x⁷/7)|₀¹ = 6π/7. (c) Now the radius of the inner circle is xⁿ, so A(x) = π⋅(1 – (xⁿ²) and V = ∫₀¹π⋅(1 – x²ⁿ) dx = π⋅(1 – x²ⁿ⁺¹/(2n+1))|₀¹ = 2nπ/(2n+1). As n → ∞, volume → π. This is the volume of the cylinder obtained by rotating the line y = 1, 0 ≤ x ≤ 1 about the x-axis. This is reasonable, because as n → ∞ the curve y = xⁿ looks more and more like the line from (0,0) to (1,0), followed by the line from (1,0) to (1,1). That is, the inner radius of the annulus approaches 0.

Suppose for 1 ≤ x ≤ 2, perpendicular to the x-axis a solid has rectangular cross-section with base x². Find the height function h(x) = x^a so the solid has volume 1

1. The value a= -2 works.

1. The base length is x² and the height is x^a, so the cross-sectional area is A(x) = base⋅altitude = x²⋅x^a = x^a+2. Then the volume is V = ∫₁²x^a+2 dx = (x^a+3)/(a + 3)|₁²= (2^a+3 – 1)/(a + 3). Then setting volume = 1 gives 2^a+3= 4 + a. This can be solved numerically, or by plotting y = 2^x+3 and y = 4 + x on the same graph. Or in this case we can guess, and find a = -2 works.