July 23, 2019, 04:08:59 am

AuthorTopic: Simpsons Rule Volume  (Read 269 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

Jefferson

• Forum Regular
• Posts: 67
• Respect: 0
Simpsons Rule Volume
« on: March 13, 2019, 10:37:36 pm »
0
Hi everyone,

I'm having trouble understanding the following solution to this question.
Why is the function value, or y, squared in this case? What exactly does that accomplishes?

Isn't volume calculated by area of cross section * height ?

With those values, wouldn't the approach be to find the area using simpson's rule (though we are not given the width of the lake, as y is the depth) and multiply it by the 'common depth not provided' of the lake?

RuiAce

• HSC Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8414
• "All models are wrong, but some are useful."
• Respect: +2289
Re: Simpsons Rule Volume
« Reply #1 on: March 14, 2019, 07:02:19 am »
+4
The cross-sectional area is not constant, unlike say for example a cylinder or a triangular pyramid. This is because we rotate the area below a curve across a certain axis.

This was the whole point of introducing volumes of solids of revolution in the 2U integration topic to begin with, and is also where we establish the formula $V =\pi \int_a^b y^2\,dx = \pi \int_a^b [f(x)]^2\,dx$ from. Have you forgotten this?

(Note that adapted here, the depth of the water not being constant is what causes the cross-sectional area to not be constant.)

Jefferson

• Forum Regular
• Posts: 67
• Respect: 0
Re: Simpsons Rule Volume
« Reply #2 on: March 14, 2019, 07:53:18 am »
0
The cross-sectional area is not constant, unlike say for example a cylinder or a triangular pyramid. This is because we rotate the area below a curve across a certain axis.

This was the whole point of introducing volumes of solids of revolution in the 2U integration topic to begin with, and is also where we establish the formula $V =\pi \int_a^b y^2\,dx = \pi \int_a^b [f(x)]^2\,dx$ from. Have you forgotten this?

(Note that adapted here, the depth of the water not being constant is what causes the cross-sectional area to not be constant.)

Ah alright.
Thanks!

jamonwindeyer

• It's Over 9000!!
• Posts: 10127
• Electrical Engineer by day, AN Enthusiast by Night
• Respect: +3027
Re: Simpsons Rule Volume
« Reply #3 on: March 14, 2019, 09:52:47 am »
+2
Ah alright.
Thanks!

Hey Jefferson! I'm more in your camp here - Using Simpson's Rule in this context is flawed because the area in this case is being linearly swept, not rotated, to produce a volume. Intuitively I think of it this way - How can we give a three dimensional measurement, with absolutely no information about the third dimension? In actual practical terms, that is nonsense, and so is the way they did this question

Rui is correct in that, when we do need to use Simpson's Rule for Volume, we just do it by approximating the volume integral using the formula. It's just that in this case it is a bit of a flawed exercise, because the volume we want isn't generated by rotation. This is a common thing that resources tend to get a bit wrong, I tended to just roll with it when I did my HSC

(An example where this works perfectly might be if we were given a vase, and radial measurements of that vase along its height, the volume integral then would be an accurate thing to do because you can think of the vase volume as a rotated area)

RuiAce

• HSC Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8414
• "All models are wrong, but some are useful."
• Respect: +2289
Re: Simpsons Rule Volume
« Reply #4 on: March 14, 2019, 10:32:19 am »
+2
To be truthful I did feel something was off, but I failed to put my head around it at the time. It’s just maths in focus being itself in my opinion.

Jefferson

• Forum Regular
• Posts: 67
• Respect: 0
Re: Simpsons Rule Volume
« Reply #5 on: March 16, 2019, 04:41:15 pm »
+1
Hey Jefferson! I'm more in your camp here - Using Simpson's Rule in this context is flawed because the area in this case is being linearly swept, not rotated, to produce a volume. Intuitively I think of it this way - How can we give a three dimensional measurement, with absolutely no information about the third dimension? In actual practical terms, that is nonsense, and so is the way they did this question

Rui is correct in that, when we do need to use Simpson's Rule for Volume, we just do it by approximating the volume integral using the formula. It's just that in this case it is a bit of a flawed exercise, because the volume we want isn't generated by rotation. This is a common thing that resources tend to get a bit wrong, I tended to just roll with it when I did my HSC

(An example where this works perfectly might be if we were given a vase, and radial measurements of that vase along its height, the volume integral then would be an accurate thing to do because you can think of the vase volume as a rotated area)

I see. That makes a lot more sense.
Thank you for clarifying!

jamonwindeyer

• It's Over 9000!!
• Posts: 10127
• Electrical Engineer by day, AN Enthusiast by Night
• Respect: +3027
Re: Simpsons Rule Volume
« Reply #6 on: March 16, 2019, 04:56:03 pm »
+1
I see. That makes a lot more sense.
Thank you for clarifying!

Very welcome, happy to help

emmajb37

• Posts: 22
• Respect: 0
Re: Simpsons Rule Volume
« Reply #7 on: March 27, 2019, 08:46:12 am »
0
Use simpson’s rule with 5 function values to find the volume of the solid formed when the curve y = ex is rotated about the y-axis from y = 3 to y = 5 , correct to 2 significant figures.

jamonwindeyer

• It's Over 9000!!
• Posts: 10127
• Electrical Engineer by day, AN Enthusiast by Night
• Respect: +3027
Re: Simpsons Rule Volume
« Reply #8 on: March 27, 2019, 10:18:30 am »
+2
Hello! The first step will be to rearrange the curve, because the area is with respect to the $y$-axis. So:
$y=e^x\implies x=\ln{y}$
Now, we need 5 function values between $y=3$ and $y=5$. So we'll do 3, 3.5, 4, 4.5, and 5. These then just get used in the Simpson's Rule formula! There are a few versions of this, here is one:
$A=\frac{h}{3}\left(f_1+f_5+4\left(f_2+f_4\right)+2f_3\right)=\frac{0.5}{3}\left(\ln{3}+\ln{5}+4\left(\ln{3.5}+\ln{4.5}\right)+2\ln{4}\right)$