Calculus Question?

Question

Calculate the volume of the solid formed when the region in the first quadrant bounded by the curve y = x^2, the axis x = 0 and the line y = 4 is rotated about the x-axis.

Answer 1

The axis x = 0 is the y-axis.
Let's use the washer method. It is a disk in which the center section is removed.

V = ∫π(R² - r²)dx = π∫(4² - (x²)²)dx

= π∫(16 - x^4)dx

= π(16x - x^5/5) | [Evaluated from 0 to 2]

= π(32 - 32/5) - 0 = 128π/5

Answer 2

The first answer gives you the volume formed when the area BELOW the curve is revolved around the x-axis. Since you want the volume formed ABOVE the curve and below line y = 4 then you must subtract that answer from the cylinder formed by rotating y = 4 around the x-axis from 0 to 2.

Answer 3

The volume is a cylinder - the volume of the rotated x^2 graph

Volume of the cylinder = pi * y^2 * x
=pi * 16 *2
=32 pi

Volume of the removed section:
limits: x = 0, x=2

volume of rotation is:
S pi * y^2 dx
where S is the integral sign

= pi * S x^4 dx
= pi * [(1/5) (x^5)] between 2 and 0
= pi * [32/5 - 0/5]

so the volume is 32 pi - 32/5 pi = 128/5 pi

Answer 4

Volume by Rotation:
V = [intergral between a and b of] ([pi]y^2) dx
a=0
b is the value of x when y=4.
y would be substituted using (y=x^2).

This will give you the area beneath the curve.

Next, find the volume by rotation for the line y=4 instead of y=x^2 (a and b are the same values as before).

The area you actually want is the difference between the two, so take your first answer from your second.

P.S.
Thanks for the challenge, it's yonks since I last did anything involving that.

Answer 5

since y=4, x = 2 is boundary.


think of it as a bunch of little disks, whose total volume you're calculating. The disks go from 0 to 2, their radiuses go from 0 to 4, and their depth is dx. Volume of such disk is

PI * r^2 * dx = PI * f(x)^2 * dx

Now integrate:

int [from 0 to 2] (PI* f(x)^2 * dx) =
PI * int [from 0 to 2] (x^4 dx) =
PI * [from 0 to 2] (1/5 * x^5 + C) =
PI * (1/5 * 2^5 + C - 1/5 * 0^5 - C) =
32/5 PI