what happens to the volume of a sphere when its diameter is doubled
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A cube with sides of 4" each contains a floating sphere with a radius of 1". What is the volume of the space outside of the sphere, inside the cube?
Possible Answers:
eleven.813 in3
64 in3
iv.187 in3
xi.813 in3
59.813 in3
Right reply:
59.813 in3
Explanation:
Volume of Cube = side3 = (four")iii = 64 in3
Volume of Sphere = (iv/three) * π * rthree = (4/3) * π * 13 = (4/3) * π * 13 = (four/3) * π = 4.187 iniii
Departure = Volume of Cube – Book of Sphere = 64 – iv.187 = 59.813 inthree
If a sphere'south diameter is doubled, by what factor is its book increased?
Caption:
The formula for the book of a sphere is 4πr3/three. Because this formula is in terms of the radius, it would be easier for us to decide how the change in the radius affects the volume. Since we are told the diameter is doubled, we need to outset determine how the alter in the diameter affects the modify in radius.
Let us call the sphere's original bore d and its original radius r. We know that d = 2r.
Let'southward call the last diameter of the sphere D. Because the diameter is doubled, we know that D = 2d. We can substitute the value of d and obtain D = 2(2r) = 4r.
Let'south phone call the last radius of the sphere R. We know that D = 2R, and so nosotros tin now substituate this into the previous equation and write 2R = 4r. If we simplify this, we see that R = 2r. This means that the final radius is twice equally large as the initial radius.
The initial volume of the sphere is 4πr3/3. The concluding volume of the sphere is 4πRthree/3. Considering R = 2r, we can substitute the value of R to obtain 4π(2r)3/three. When we simplify this, nosotros get 4π(8r3)/iii = 32πr3/3.
In gild to determine the factor by which the volume has increased, we need to find the ratio of the final book to the initial book.
32πr3/3 divided past 4πr3/3 = 8
The volume has increased past a factor of 8.
A sphere increases in volume past a factor of 8. By what factor does the radius change?
Explanation:
Book of a sphere is 4Πr3/three. Setting that equation equal to the original book, the new volume is given as 8*4Πr3/3, which tin be rewritten as 4Π8r3/three, and tin be added to the radius value by 4Π(2r)three/iii since 8 si the cube of two. This ways the radius goes upwardly by a factor of 2
A foam ball has a volume of two units and has a diameter of x. If a second foam ball has a radius of 2x, what is its volume?
Possible Answers:
sixteen units
128 units
four units
2 units
8 units
Explanation:
Careful non to mix up radius and bore. First, nosotros need to place that the second ball has a radius that is iv times every bit large as the first ball. The radius of the showtime brawl is (1/2)x and the radius of the 2nd ball is 2x. The volume of the 2d ball will be 43, or 64 times bigger than the outset ball. And then the 2nd brawl has a volume of 2 * 64 = 128.
A cross-section is made at the eye of a sphere. The expanse of this cross-section is 225 π square units. How many cubic units is the total book of the sphere?
Possible Answers:
1687.5π
4500π
None of the other answers
13500π
3375π
Explanation:
The solution to this is unproblematic, though just have it stride-by-pace. Kickoff observe the radius of the circular cross-department. This will requite u.s. the radius of the sphere (since this cross-section is at the center of the sphere). If the cantankerous-section has an area of 225π, nosotros know its area is defined by:
A = 225π = πr 2
Solving for r, we get r = 15.
From here, nosotros merely need to use our formula for the book of a sphere:
V = (4 / 3)πr three
For our data this is: (4 / iii)π * xviii = 4π * xv2 * 5 = 4500π
At x = 3, the line y = 4x + 12 intersects the surface of a sphere that passes through the xy-airplane. The sphere is centered at the point at which the line passes through the x-axis. What is the volume, in cubic units, of the sphere?
Possible Answers:
4896π
4896π√(17)
2040π√(vii)
816π√(eleven)
None of the other answers
Correct answer:
4896π√(17)
Caption:
We demand to ascertain 2 values: The center point and the signal of intersection with the surface. Allow'due south do that kickoff:
The center is defined by the x-intercept. To discover that, gear up the line equation equal to 0 (y= 0 at the 10-intercept):
0 = 4x + 12; ivx = –12; x = –3; Therefore, the center is at (–3,0)
Next, we need to discover the bespeak at which the line intersects with the sphere's surface. To practise this, solve for the point with x-coordinate at 3:
y = 4 * three + 12; y = 12 + 12; y = 24; therefore, the point of intersection is at (3,24)
Reviewing our information so far, this means that the radius of the sphere runs from the heart, (–3,0), to the edge, (3,24). If we observe the distance between these ii points, nosotros can ascertain the length of the radius. From that, nosotros will be able to calculate the volume of the sphere.
The distance between these 2 points is defined by the distance formula:
d = √( (x 1 –x 0)ii + (y ane –y 0)ii)
For our information, that is:
√( (3 + 3)2 + (24 – 0)2) = √( half dozen2 + 242) = √(36 + 576) = √612 = √(2 * ii * three * three * 17) = 6√(17)
Now, the volume of a sphere is defined past:V = (four/3)πr 3
For our data, that would exist: (4/3)π * (6√(17))three = (4/3) * six3 * 17√(17) *π = four * ii * 6ii * 17√(17) *π = 4896π√(17)
The surface area of a sphere is 676 π in2. How many cubic inches is the volume of the aforementioned sphere?
Possible Answers:
(2197π)/three
8788π
(8788π)/3
(2028π)/3
2028π
Explanation:
To brainstorm, we must solve for the radius of our sphere. To exercise this, retrieve the equation for the surface surface area of a sphere: A = 4 πr 2
For our information, that is: 676 π = iv πr ii ; 169 = r two; r = xiii
From this, information technology is easy to solve for the volume of the sphere. Recall the equation:
V = (iv/three)πr iii
For our data, this is: V = (iv/3)π * thirteen3 = (4π * 2197)/3 = (8788π)/iii
What is the deviation between the volume and surface expanse of a sphere with a radius of half dozen?
Explanation:
Surface area = fourπr 2 = four * π * 62 = 144π
Volume = 4πr 3/iii = 4 * π * 63 / 3 = 288π
Book – Surface Expanse = 288π – 144π = 144π
A sphere is perfectly contained within a cube that has a volume of 216 units. What is the book of the sphere?
Correct answer:
Explanation:
To begin, we must make up one's mind the dimensions of the cube. This is done by solving the elementary equation: 
Nosotros know the volume is 216, allowing u.s. to solve for the length of a side of the cube.
Taking the cube root of both sides, nosotros go s = half-dozen.
The diameter of the sphere volition be equal to side of the cube, since the question states that the sphere is perfectly contained. The diameter of the sphere will be 6, and the radius will be three.
Nosotros can plug this into the equation for volume of a sphere:
We can cancel out the 3 in the denominator.
Simplify.
The surface expanse of a sphere is 
Correct reply:
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