NB : Show your working

1) Give me the formula for the volume V of a solid formed by taking a sphere of radius R and boring a cylindrical hole of radius r through it with the centre of the sphere lying on the axis of symmetry of the cylinder.

2) Find the mass of the solid sphere of radius 5 metres whose variable density at every point on the sphere is such that it is linearly proportion to the distance from the centre of the sphere.

3) Give me the coordinates of the centre of the region in the plane bounded by the graphs of f(x) = 4 - x² and g(x) = x + 2 ?

4) I placed the following six masses : 7kg, 6kg, 5kg, 4kg, 3kg and 2kg on the cartesian plane and they have the following coordinates (2, 3), ( ½, 4), (-3, 2), (-4, 1), (3, - 2) & (1, 1) respectively. What are the coordinates of the point on which I can balance the cartesian plane on the tip on a pencil without it tipping over?

Martinez39:
NB : Show your working

1) Give me the formula for the volume V of a solid formed by taking a sphere of radius R and boring a cylindrical hole of radius r through it with the centre of the sphere lying on the axis of symmetry of the cylinder.

2) Find the mass of the solid sphere whose centre is at the origin but whose variable density is such that it is linearly proportion to the distance from it's centre.

3) Give me the coordinates of the centre of the region in the plane bounded by the graphs of f(x) = 4 - x² and g(x) = x + 2 ?

4) I placed the following six masses : 7kg, 6kg, 5kg, 4kg, 3kg and 2kg on the cartesian plane and they have the masses following coordinates (2, 3), ( ½, 4), (-3, 2), (-4, 1), (3, - 2) & (1, 1) respectively. What are the coordinates of the point on which I can balance the cartesian plane on the tip on a pencil without it tipping over?

Can you explain number 4 a bit further?

Darivie04:


Can you explain number 4 a bit further?

When you place various masses on a plane, there is a point on the plane at which the plane(together with the masses) would balance on a pencil tip. What is that centre of balance for the mass system I gave in question four?

Will be dropping solutions for 1,3 and 4 later. Number 2 still seems a bit confusing though..I would need some explanations on that.

Darivie04:
Will be dropping solutions for 1,3 and 4 later. Number 2 still seems a bit confusing though..I would need some explanations on that.

Thanks. Number 2 was incomplete. I have modified it.

@Nurrex

What's wrong? Why are you posting closed posts?

Mathematicians no dey Nairaland?

Sorry oh! I've been busy.

Martinez39:


4) I placed the following six masses : 7kg, 6kg, 5kg, 4kg, 3kg and 2kg on the cartesian plane and they have the following coordinates (2, 3), ( ½, 4), (-3, 2), (-4, 1), (3, - 2) & (1, 1) respectively. What are the coordinates of the point on which I can balance the cartesian plane on the tip on a pencil without it tipping over?

-0.1111, 2.03737
I will upload workings later.

I am on android and can't type comfortably as I would have loved but let me try.
Just start from coordinate 0,0 and treat the x and y coordinates as the x and y magnitude of the weights. This will require us to multiply the x and y value by the weight attached to each point.That gives the resultant x and y axis weight component of all loads.

Next step is to add them all together ( all x axis added together separately from y axis )

finally divide by the net weight of loads available.

Martinez39:
NB : Show your workings :
3) Give me the coordinates of the centre of the region in the plane bounded by the graphs of f(x) = 4 - x² and g(x) = x + 2 ?

Find the inspection of the two equations:
4-X² = x+2
gives
X²+X-2=0 and this gives
X= -2 and 1

substituting these values in the equations gives values for y as 0 and 3. This corresponds to point (1,3) and (-2,0)
The centre of the equation should be answer and is (-0.5, 1.5)

Please respond and tell if I got any right.

CodeTemplar:

Find the inspection of the two equations:
4-X² = x+2
gives
X²+X-2=0 and this gives
X= -2 and 1

substituting these values in the equations gives values for y as 0 and 3. This corresponds to point (1,3) and (-2,0)
The centre of the equation should be answer and is (-0.5, 1.5)

Please respond and tell if I got any right.

Its wrong, you can plot it and check.

You're supposed to find center of the region bounded.

CodeTemplar:

-0.1111, 2.03737
I will upload workings later.

I am on android and can't type comfortably as I would have loved but let me try.
Just start from coordinate 0,0 and treat the x and y coordinates as the x and y magnitude of the weights. This will require us to multiply the x and y value by the weight attached to each point.That gives the resultant x and y axis weight component of all loads.

Next step is to add them all together ( all x axis added together separately from y axis )

finally divide by the net weight of loads available.

The sum of moments about the y-axis is:
Σ(m_ix_i) = (7 × 2) + (6 × ½) + (5 × -3) + (4 × -4) + (3 × -2) + (2 × 1) = -18
The sum of moments about the x-axis is:
Σ(m_iy_i) = (7 × 3) + (6 × 4) + (5 × 2) + (4 × 1) + (3 × -2) + (2 × 1) = 55
The sum of masses
Σm_i = 7 + 6 + 5 + 4 + 3 + 2 = 27.

The point of balance is (a, b), where
a = Σ(m_ix_i) ÷ Σm_i = -2/3
b = Σ(m_iy_i) ÷ Σm_i = 55/27

Point of balance is (-2/3 , 55/27).

Martinez39:
The sum of moments about the y-axis is:
Σ(m_ix_i) = (7 × 2) + (6 × ½) + (5 × -3) + (4 × -4) + (3 × -2) + (2 × 1) = -18
The sum of moments about the x-axis is:
Σ(m_iy_i) = (7 × 3) + (6 × 4) + (5 × 2) + (4 × 1) + (3 × -2) + (2 × 1) = 55
The sum of masses
Σm_i = 7 + 6 + 5 + 4 + 3 + 2 = 27.

The point of balance is (a, b), where
a = Σ(m_ix_i) ÷ Σm_i = -2/3
b = Σ(m_iy_i) ÷ Σm_i = 55/27

Point of balance is (-2/3 , 55/27).

I think I made a mistake in the x-axis then. I used MS Excel thinking I could copy it into Nairaland for better formatting but, as you can see...
I got -3 in the part where you got -18.

Darivie04:

Its wrong, you can plot it and check.

You're supposed to find center of the region bounded.

I understand you now. I was working with the assumption that the equation of the line connecting the two intersections is their middle. I think I need sum both equations and divide by two to get and equation of their average at at points. I will try that.