Calculus: Volumes! Can you PLEASE help me with these?

Question

Find the Volume of the solid described. Give exact answers as reduced fractions in terms of pi.

1. The base is a circle with radius a. Find the volume if cross sections perpendicular are squares.

2. The base is the region enclosed by a circle of radius 7cm: find the volume if cross sections perpendicular to a fixed diameter are equilateral triangles.

3. Base is region bounded by y=1/x, y=0, x=1, x=4 and cross sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the xy-plane.

4. The base of a solid is bounded by x=y^2 and x=4. Find the volume if cross sections are:
a) perp. to -axis and are isosceles triangles with base in xy-plane and height is 1/2 of base.
b) perp. to x-axis and are ososceles right triangles with one =~ side in the xy-plane.
c) perp. to y-axis and are squares with one side in the xy-plane.
d) perp. to y-axis and are semicircles with diameter on xy-plane.

please please help me...i am pleading. you have no idea. help... :(

Answer 1

The key here is to try and determine what the solid described is. If it's not a classical solid shape (cylinder, prism, etc), then you have to use the calculus method of V = ∫A (h) dh where:

h = is any dimension of the figure,
A(h) = area of the cross-sections perpendicular to h described as a function of the position along h

1.) The solid described is a cylinder with height 2a. (because the height = diameter)
Vcyl = πr^2h
Vcyl = πa^2(2a)
Vcyl = 2πa^3

2.) The solid described is a cone with a height that is twice the radius (14 cm).
Vcone = 1/3 bh
b = πr^2
Vcone = 1/3 * π * (7 cm)^2 * (14cm)
V = 228 1/3 * π

3.) Use V = ∫A (h) dh to find the height.
Step 1: Find area of base, so V = 1/3 bh
A = ∫(4, 1) 1/x dx
A = ln |x| | (4, 1)
A = ln 4 - ln 1
A = ln 4 (because ln 1 = 0)

Step 2: Define height.
The height of the structure is defined as the altitude of the isoceles right triangle defined in the cross section. That altitude is defined by h = a sin 45, because an isoceles right triangle has an angle of 45 degrees. a is the length of a non-hypotenuse side.

c^2 = a^2 + a^2 (since the legs are same size.
c^2 = 2a^2
c = 1/x
1/x^2 = 2a^2
1/2x^2 = a^2
√2/2 * x = a
h = √2/2 * x * sin 45
h = √2/2 * x * √2/2
h = 1/2x

Put it all together and integrate:
V = ∫A (h) dh
V = ∫ln 4 * 1/2 x dx
V = 1/2 ln 4 ∫(4,1) x dx
V = 1/2 ln 4 * x^2/2 | (4,1)

Note: 1/2 ln 4 = ln √4 = ln 2

V = ln 2 * (4^2/2 - 1^2/2) = ln 2 (8 - 1/2) = 7.5 ln 2

This is the same method you'll use for #4.