Notes PDF
More Challenging Problems
1. Find the volume of the solid with right isosceles triangular cross-section perpendicular to the x-axis, with base x2, for 0 ≤ x ≤ 1
Answer
1. Volume = 1/10.
Solution
1. A right isosceles triangle with base x2 has altitude x2 and so area A(x) = (1/2)⋅base⋅altitude (1/2)⋅x2⋅x2 = (1/2)⋅x4 Then the volume is: V = ∫01A(x) dx = ∫01(1/2)⋅x4dx = x5/10|01= 1/10.
2. Find the volume if the solid with elliptical cross-section perpendicular to the x-axis, with semi-major axis x2 and semi-minor axis x3, for 0 ≤ x ≤ 1
Answer
1. Volume = π/6.
Solution
1. Recall an ellipse with semi-major axis a and semi-minor axis b has area πab, so this ellipse with semi-major axis x2 and semi-minor axis x3 has the area: A(x) = π⋅x2⋅x3 = π⋅x5. Then the volume is V= ∫01A(x) dx = ∫01π⋅x5dx = π⋅x6/6|01 = π/6.
3. Find the volume of the solid with circular cross-section of radius cos3/2(x), for 0 ≤ x ≤ π/2.
Answer
1. Volume = 2π/3.
Solutions
1. The area is A(x) = π(cos3/2(x)2 =πcos3(x). The volume is V = ∫0π/2A(x) dx = ∫0π/2πcos3(x) dx= π∫0π/2cos(x)(1 – sin2(x)) dx = π∫0π/2cos(x)dx – π∫0π/2cos(x)sin2(x) dx = πsin(x)|0π/2 – πsin3(x)/3= |0π/2 = π – π/3 = 2π/3. where cos(x)sin2(x) is integrated using the substitution u = sin(x), so du = cos(x) dx.
4. Find the volume of the solid with cross-section a rectangle of base x and height ex2
Answer
1. Volume= (e – 1)/2.
Solution
1. The area is A(x) = base ⋅ height = x⋅ex2. The volume is V = ∫01A(x) dx = ∫01 x⋅ex2 dx= (ex2)/2|01 = (e – 1)/2. where x⋅ex2 was integrated using the substitution u = x2, so du = 2xdx.
5. Find the volume of the solid obtained by rotating the curve y = x2, -1 ≤ x ≤ 1, about the x-axis.
Answer
1. Volume = 2π/5.
Solution
1. The cross-sections are circles of radius x2, so the cross-sectional area is A(x) π⋅(x2)2π⋅x4 The volume is V = ∫-11A(x) dx = ∫-11 π⋅x4 dx = π⋅(x5/5)|-11 = 2π/5