[SOLVED] finding the volume of a beaker

**ogglock** · February 20th 2010, 12:38 PM

i need to find the volume of this beaker

http://img13.imageshack.us/img13/1264/72680665.png

other than the information given in the picture, you are also told that the diameter of the opening is 1.2. I could not figure out the radius of the spherical part to find the total volume of the beaker

**earboth** · February 20th 2010, 11:27 PM

Originally Posted by ogglock

i need to find the volume of this beaker

http://img13.imageshack.us/img13/1264/72680665.png

other than the information given in the picture, you are also told that the diameter of the opening is 1.2. I could not figure out the radius of the spherical part to find the total volume of the beaker

1. Draw a sketch of the cross-section. (see attachment)

2. You are dealing with a right triangle. (Coloured in grey)

3. According to Euclid's theorem you have:

$h \cdot (2r-h) = 0.6^2$

4. According to your sketch $(2r - h) = 7.3\ cm~\implies~r=\frac12 (7.3+h)$ . Therefore the equation at 3. becomes:

$h \cdot 7.3 = 0.6^2$

5. Solve for h, determine r.

**ogglock** · February 20th 2010, 11:31 PM

i appreciate the help, but is there an easy way to find this? i have not learned how to use Euclid's theorem that you just used.

**earboth** · February 20th 2010, 11:58 PM

Originally Posted by ogglock

i appreciate the help, but is there an easy way to find this? i have not learned how to use Euclid's theorem that you just used.

I don't know if it is easier ...

Use the right triangle at the top of the grey triangle. Use Pythagorean theorem:

$0.6^2+(r-h)^2 = r^2$

$0.6^2= r^2 -(r-h)^2$

$0.6^2= r^2 -r^2+2rh-h^2$

$0.6^2= 2rh-h^2$

$0.6^2= h(2r-h)$

Plug in (2r-h) = 7.3 and solve for h, determine r.

**Peleus** · February 21st 2010, 03:46 AM

A rough guide would be -

7.3cm high for the spherical part, hence a radius of 4.65cm.

Volume of the bottom is

$<br /> \frac{4}{3}\pi r^3<br />$
$<br /> \frac{4}{3}\pi 4.65^3<br />$
= 421.16cm^3 for the sphere.

Finding the volume of the cylindrical top assuming a radius of 0.6cm and a height of 4.7cm. This gives ~5.32cm^3

Total volume around 426.48cm^3. This doesn't take into account inaccuracies such as doubling up on the joint where the sphere and cylinder meet etc, but it's hardly an accurate drawing to work off to begin with.

**earboth** · February 21st 2010, 05:04 AM

Originally Posted by Peleus

7.3cm high for the spherical part, hence a radius of 4.65cm.

...

Obviously you made a tiny but very effective mistake: From a diameter of 7.3 cm you'll get a radius of r = 3.65 cm.

**Archie Meade** · February 21st 2010, 05:51 AM

Here is the problem,
and the reason for earboth's geometry.

**ogglock** · February 21st 2010, 12:20 PM

thank you to everbody who helped me with this question i appreciate it. Sorry to be more of a pain but i got a couple of surface area questions that i need help with.

Question 1:

A 21.3 cm diameter hole is drilled in a cube with 48.4 cm long sides as shown here (sketch is not to scale). Determine the total surface area of this shape in square centimeters.
Express your answer to three significant digits.

Reference Pic:http://img404.imageshack.us/img404/9702/84313854.png

Question 2:

A tank used to transport liquids is made up of a cylinder with semi-spheres on its ends as shown (drawing not to scale). The length of the cylindrical portion of the tank is 2.2 m and its height is 2.1 m. What is the surface area of the outside of the tank in square metres?

Reference Pic: http://img710.imageshack.us/img710/1430/83625107.png

I did the second question and got 49.8. I wanted to know if this was the right answer, if it is not how do i get to the right answer.

thanks in advance

**earboth** · February 22nd 2010, 06:41 AM

Originally Posted by ogglock

...

Question 1:

A 21.3 cm diameter hole is drilled in a cube with 48.4 cm long sides as shown here (sketch is not to scale). Determine the total surface area of this shape in square centimeters.
Express your answer to three significant digits.

Reference Pic:http://img404.imageshack.us/img404/9702/84313854.png

Question 2:

A tank used to transport liquids is made up of a cylinder with semi-spheres on its ends as shown (drawing not to scale). The length of the cylindrical portion of the tank is 2.2 m and its height is 2.1 m. What is the surface area of the outside of the tank in square metres?

Reference Pic: http://img710.imageshack.us/img710/1430/83625107.png

I did the second question and got 49.8. I wanted to know if this was the right answer, if it is not how do i get to the right answer.

thanks in advance

1. If you have a new question please start a new thread. Otherwise nobody will see that you need some additional support.

to Q1: The total surface of this solid consists of:

A = (surface of cube) - 2(base circles of cylinder) + (curved surface of cylinder)

Let s denote the length of the edge of the cube and r the radius of the base circle of the cylinder which have of course the height s then you get:

$A = 6 \cdot s^2 - 2 \cdot \pi r^2 + 2\pi \cdot r \cdot s$

Plug in the known values!

to Q2: I'm very interested to learn how you got this result.

The radius of the cylinder and the sphere is r = 1.05 m, the length of the cylinder is l = 2.2 m.

The the surface area consists of the curved surface area of the cylinder and the surface of a complete sphere:

$A = \underbrace{2\pi \cdot r \cdot l}_{\text{curved surface of cylinder}} + \underbrace{4\pi \cdot r^2}_{\text{surface of sphere}}$

Plug in the values you already know. I've got A = 28.369 m²

**ogglock** · February 22nd 2010, 11:07 AM

Originally Posted by earboth

1. If you have a new question please start a new thread. Otherwise nobody will see that you need some additional support.

to Q1: The total surface of this solid consists of:

A = (surface of cube) - 2(base circles of cylinder) + (curved surface of cylinder)

Let s denote the length of the edge of the cube and r the radius of the base circle of the cylinder which have of course the height s then you get:

$A = 6 \cdot s^2 - 2 \cdot \pi r^2 + 2\pi \cdot r \cdot s$

Plug in the known values!

to Q2: I'm very interested to learn how you got this result.

The radius of the cylinder and the sphere is r = 1.05 m, the length of the cylinder is l = 2.2 m.

The the surface area consists of the curved surface area of the cylinder and the surface of a complete sphere:

$A = \underbrace{2\pi \cdot r \cdot l}_{\text{curved surface of cylinder}} + \underbrace{4\pi \cdot r^2}_{\text{surface of sphere}}$

Plug in the values you already know. I've got A = 28.369 m²

I see what you did for the second question. what i did for it was take the SA of the cylinder which i found was (2pi r^2)+ (2pi r h) and added it to the SA of the hemisphere multiplied by 2

**earboth** · February 22nd 2010, 12:01 PM

Originally Posted by ogglock

I see what you did for the second question. what i did for it was take the SA of the cylinder which i found was (2pi r^2)+ (2pi r h) and added it to the SA of the hemisphere multiplied by 2

Probably you see now that the tank is constructed by
- the curved surface of the cylinder that means without the circle at the base and at the top.

- and the surface of a complete sphere that means without the circle at the base.

If you use the formulas for the surface area of a complete cylinder and a complete semi-sphere then your result must be too large.

Thread: [SOLVED] finding the volume of a beaker

LinkBack

Thread Tools

Display

[SOLVED] finding the volume of a beaker

Similar Math Help Forum Discussions

[SOLVED] Theoretical Proof for volume of a cone in relation to the volume of a cylind

[SOLVED] Finding volume with integration

[SOLVED] Finding the volume of this figure

finding volume help!!!

finding the volume

Search Tags