Calculus...Volumes of Solids?

Question

A ball of radius 17 has a round hole of radius 6 drilled through its center. Find the volume of the resulting solid.

(Subtracting one from the other is incorrect)

V = integral (low: a, up: b) A(x) dx

Answer 1

Volume of a spherical bead

A ball of radius 17 has a round hole of radius 6 drilled through its center. Find the volume of the resulting solid.
We will use the washer method for a volume of rotation. Consider the hole to be drilled along the x axis. Since the ends are gone, we must recalculate the endpoints as the limits of integration.

6^2 + X^2 = 17^2

36 + X^2 = 289

X^2 = 253

X = +/- sqrt(253)

The area of the washer is the difference of the areas of the outer and inner circles.
The outer circle has radius sqrt(289 - x^2).
The inner circle has radius 6.

A(x) = pi sqrt(289 - x^2)^2 - pi 6^2

A(x) = pi (289 - x^2) - pi 36

A(x) = pi (253 - x^2)

V = integral (low: -sqrt(253), up: sqrt(253)) pi (253 - x^2) dx

Factor the pi outside the integral.

V = pi integral (low: -sqrt(253), up: sqrt(253)) (253 - x^2) dx

253 - x^2 is an even function, so we can double the integral on the positive side starting from 0. This reduces the opportunity to make an error during the evaluation step.

V = 2 pi integral (low: 0, up: sqrt(253)) (253 - x^2) dx

Integrate.

V = 2 pi evaluate (low: 0, up: sqrt(253)) (253 x - x^3 / 3)

Evaluate.

V = 2 pi (253 sqrt(253) - sqrt(253)^3 / 3)

Gather like terms and simplify.

V = 2 pi (253 sqrt(253) - 253 sqrt(253) / 3)

V = 2 pi 253 sqrt(253) (1 - 1/3)

V = 2 pi 253 sqrt(253) 2/3

V = 4 pi 253 sqrt(253) / 3

V = 1012 pi sqrt(253) / 3

Answer 2

Let h = height from the center of the cylindrical hole. Then the volume integral is:

V = 2 ∫ (0 to √(17² - 6²)) (π(17² - h² - 6²)) dh = (π/3)1012√253

or about 16,856.577

Answer 3

Equation of the circle,

7^2 = x^2 + y^2
since there is a hole inside the ball with radius 6. Then your limit of integration will be,

y = sqrt( 17^2 - 6^2)
y = 15.9
limits of integration are -15.9 to 15.9

V =∫∏ x^2dy - ∏6^2(15.9)

Solve this one it will result,
V = 18654.82

Answer 4

Wow you are REALLY trying to stump someone huh. Hah. Well i can tell you how to solve it but someone else will have to do the math. Lets see what we know:

Height of ball and "cylinder", 34.
The "cylinder" has round sides tho.
Width of "Cylinder" is 12.

What we need now is arc length formula. Find the height of the arc, use that to find the area of the circular section of the sphere. Multiply that by 2 = x. Take the height of the arcs and multiply by 2 and subtract that from 34 = y.
12y + x = z = Volume of the "Cylinder"
Subtract z from the volume of the ball (907.9)
And there you have it. Volume of the solid. i wish i knew the formulas =\

Answer 5

It is easier to calculate half of volume of hole and subtract twice of it from the volume of sphere.
therefore
volume = (4/3)*pi*(17^3) - 2*(half of volume of hole)

half of volume of hole can be obtained by integrating on a circle with radius of 6 that bounds from plane z=0 to the surface of sphere.
Now it is covenient to consider an annulus with thickness of dr and radius of r and integrate from 0 to 6. The volume of the annulus bounded between plane z=0 and the surface of sphere is height* area of annulus=height* (2*pi*r*dr)
from a right triangle, height can be written in terms of r as
square root of (17^2-r^2) therefore
half of volume of hole=∫ (2*pi*r)*square root ((289-r^2)) dr from 0 to 6=
(-2*pi/3)*(289-r^2)^(3/2) with r from 0 to 6
the rest is easy to derive.

Answer 6

Bowling Ball