How to find the volume of a quarter-sphere?
1 answer:
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1- Solution using graphs:
Take a look at the attached images.
The red graph represents the first given function while the blue graph represents the second given function.
We can note that the two graphs are the same line (they overlap).
This means that any chosen point on one of them will satisfy the other.
This means that there are infinite number of solutions to these two equations.
2- Solution using substitution:
The first given equation is:
y = -5x + 3 ...........> equation I
The second given equation is:
2y + 10x = 6 ...........> equation II
Substitute with I in II and solve as follows:
2(-5x+3) + 10x = 6
-10x + 6 + 10x = 6
0 = 0
This means that there are infinitely many solutions to the given system of equations.
Hope this helps :)
Take a look at the attached images.
The red graph represents the first given function while the blue graph represents the second given function.
We can note that the two graphs are the same line (they overlap).
This means that any chosen point on one of them will satisfy the other.
This means that there are infinite number of solutions to these two equations.
2- Solution using substitution:
The first given equation is:
y = -5x + 3 ...........> equation I
The second given equation is:
2y + 10x = 6 ...........> equation II
Substitute with I in II and solve as follows:
2(-5x+3) + 10x = 6
-10x + 6 + 10x = 6
0 = 0
This means that there are infinitely many solutions to the given system of equations.
Hope this helps :)
Problem 2
Part (a)
The 3D shape formed when rotating around the y axis forms a pencil tip
The shape formed when rotating around the x axis is a truncated cone turned on its side.
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Part (b)
Check out the two diagrams below.
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Problem 3
Answer: Choice A and Choice C
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Explanation:
Think of stacks of coins. Let's say we had 2 stacks of 10 quarters each. The quarters are identical, so they must produce identical volumes. Those sub-volumes then add up to the same volume for each stack. Now imagine one stack is perfectly aligned and the other stack is a bit crooked. Has the volume changed for the crooked stack? No, it hasn't. We're still dealing with the same amount of coins and they yield the same volume.
For more information, check out Cavalieri's Principle.
With all that in mind, this leads us to choice C. If the bases are the same, and so are the heights, then we must be dealing with the same volumes.
On the other hand, if one base is wider (while the heights are still equal) then the wider based block is going to have more volume. This leads us to choice A.
