stewart, 8th ed, problem 54

EJE12345

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Stewart, Early Transcendentals, 6.2, problem 54: find volume of solid - base S is a circular disc with radius R. Parallel cross-sections perpendicular to base are squares. I did it 2 ways, and got different answers. First one goes along with the chapter examples, and I think is fine. Second one (integrating with respect to angle theta) seems like it should work, but different answer. Where do I go wrong? thanks
 

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EJE12345

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need help with problem from Stewart's Early Transcendental's, 8th ed, 6.2.54. find volume of solid: base is disk with radius r, parallel cross-sections perpendicular to base squares. i have two solutions: first follows along with the ideas in the examples of the section, finding area of square as a function of point x on the circle, then integrating over 0 to the radius. the second solution i think it wrong but can't tell why - looking at the the area of the square as a function of the angle theta defined in the image, then integrating from 0 to pi/2. where am i going wrong? thanks
1618698219730.png
 

nasi112

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Do you have the correct answer?

Is the answer in the book \(\displaystyle \frac{16}{3}r^3\)?
 

EJE12345

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no book answer, even number problem. im pretty sure it is (16/3)R^3; i want to know why the second method is incorrrect, and how to make it correct
 

nasi112

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First, you have to be sure your first method is correct, then we can go and check the second method.

I think that your first method is wrong. What do you say?
 

EJE12345

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i am happy to hear why it is incorrect. the solid created is unusual, but really just a bunch of square slices put together. the sides of the squares vary with the radius of the circle. there could be a flaw in the set up certainly.
 

nasi112

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Your final answer is correct by coincidence because we are moving in a circle, but setting the integral is wrong.

if we let \(\displaystyle 2x\) be one side of the square, then its area is \(\displaystyle 4x^2\), then we have to take the height as \(\displaystyle dy\), not \(\displaystyle dx\), so the integral is

\(\displaystyle V = \int_{-r}^{r} 4x^2 \ dy = 8\int_{0}^{r} r^2 - y^2 \ dy\)
 

EJE12345

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i see, i got terms mixed up when transposing from a note book. so you agree the first answer is correct.

do you know why the method is incorrect, using trig functions instead?

i was a math major in college 20y ago, im relearning a lot of this just as a hobby, so i dont have a teacher/professor. the answer to my question may be apparent later as i go through the book. thanks for the help.
 

Dr.Peterson

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Stewart, Early Transcendentals, 6.2, problem 54: find volume of solid - base S is a circular disc with radius R. Parallel cross-sections perpendicular to base are squares. I did it 2 ways, and got different answers. First one goes along with the chapter examples, and I think is fine. Second one (integrating with respect to angle theta) seems like it should work, but different answer. Where do I go wrong? thanks
1618710676150.png
The element of volume is not given by \(A(\theta)d\theta\); the thickness of a slice is not \(d\theta\).

Have you learned anything about integrating in polar coordinates?
 

EJE12345

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thanks a bunch
 

EJE12345

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not yet... i cant recall the details of change of coordinates (math major 20y ago, now revisiting stuff) - when i get to polar coordinates in the book ill revisit this.
thanks
 

skeeter

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\(\displaystyle x = r\cos{\theta} \implies dx = -r\sin{\theta} \, d\theta\)

\(\displaystyle dA = (2r\sin{\theta})^2\)

\(\displaystyle V = \int_{\pi}^0 (4r^2\sin^2{\theta})(-r\sin{\theta}) \, d\theta\)

note ... rotating \(\displaystyle \theta \text{ from } \pi \to 0 \implies dx > 0\)

\(\displaystyle V = 4r^3 \int_{\pi}^0 (1-\cos^2{\theta})(-\sin{\theta}) \, d\theta\)

\(\displaystyle u = \cos{\theta} \implies du = -\sin{\theta} \, d\theta\)

\(\displaystyle V = 4r^3 \int_{-1}^1 (1-u^2) \, du = 4r^3 \bigg[u - \frac{u^3}{3}\bigg]_{-1}^1 = 4r^3\bigg[\frac{2}{3} - \left(-\frac{2}{3}\right)\bigg] = \frac{16r^3}{3}\)

cross_section_Vol.jpg
 

nasi112

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i see, i got terms mixed up when transposing from a note book. so you agree the first answer is correct.

do you know why the method is incorrect, using trig functions instead?

i was a math major in college 20y ago, im relearning a lot of this just as a hobby, so i dont have a teacher/professor. the answer to my question may be apparent later as i go through the book. thanks for the help.
First, I apologize to you because I was wrong. Your set up for the first integral is correct. I have just realized that you have used \(\displaystyle 2y\) for one side of the square instead of \(\displaystyle 2x\) which is also correct.

To do this in trigonometry is very easy. You can follow skeeter way, or you can continue with my integral.

Last time, we had this integral,

\(\displaystyle V = \int_{-r}^{r} 4x^2 \ dy = 8\int_{0}^{r} r^2 - y^2 \ dy\)

Change the variables to polar coordinate.

\(\displaystyle y = r\sin\theta\)
\(\displaystyle dy = r\cos\theta \ d\theta\)

The integral will become,

\(\displaystyle V = 8\int_{0}^{\pi/2} (r^2 - r^2\sin^2 \theta) (r \cos \theta) \ d\theta = \frac{16r^3}{3}\)

Can you figure out why we chose the limit of integration as \(\displaystyle 0\leq \theta \leq \frac{\pi}{2}\)?
 

EJE12345

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thanks guys for doing that. i do recall change of variables from an analysis course 20y ago... looks like this is a few chapters ahead in the book
 
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