What's Next in Calculus: Solving Volumes with Integrals

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SUMMARY

The discussion focuses on solving volume problems using integrals in calculus, specifically the integral of the function y = 1/x between x = 1 and x = 2. The correct setup for the volume integral is identified as \int^{1}_{2} \pi [\frac{1}{x}]^2 dx, emphasizing the importance of using the formula for the volume of a solid of revolution. Participants clarify the antiderivative process, highlighting the power law integral and correcting errors in the calculations. The conversation concludes with a successful resolution of the problem, reinforcing the understanding of antiderivatives in this context.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with the concept of volumes of solids of revolution
  • Knowledge of power law integrals
  • Ability to compute antiderivatives
NEXT STEPS
  • Study the method of disks and washers for calculating volumes
  • Practice solving power law integrals in various contexts
  • Review the concept of antiderivatives and their applications in calculus
  • Explore examples of volume calculations in calculus textbooks
USEFUL FOR

Students studying calculus, educators teaching integral calculus, and anyone looking to improve their understanding of volume calculations using integrals.

physicsed
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[SOLVED] volumes... last week of calc

i was absent when we went over the volume section.

y= 1/x, x= 1, x= 2, y= 0,; about the x-axis

<br /> \int^{1}_{2} \pi \frac{1}{x} dx<br />

i don't know what's next. can anyone inform me, please
 
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You could actually do the integration, but that would be wrong because you haven't set it up right either. You integrate pi*r^2 where r is the radius of the disk over the volume. I'd suggest checking a few examples in your textbook.
 
\int^{1}_{2} \pi [\frac{1}{x}]^2 dx

if you help me with this problem, i am sure i will get the rest of the problems.
 
It's a power law integral. Like x^n. What's n in this case? What's the antiderivative?
 
\pi \int^{2}_{1} \frac{1}{x^2} dx
 
Last edited:
Fine start. Now what's the antiderivative of 1/x^2?
 
\pi [ \frac{x^{-3}}{-3}]
 
Beep. Wrong. The antiderivative of x^n is x^(n+1)/(n+1). What's n in this case? Unless that's a careless error because you are paying more attention to texing than thinking, you may have missed more than 'volumes'.
 
Last edited:
since it's divided by 1, isn't it negative n
=x^-2
 
  • #10
Yes, it is. Can you fix your antiderivative?
 
  • #11
(-1/3)x^-3
 
  • #12
Beep. Beep. Beep. You goofed it again. What's -2+1? Think this time.
 
  • #13
righttt

pi [(1/-1)x^-1]
 
  • #14
physicsed said:
righttt

pi [(1/-1)x^-1]

Much better! You've got it now, right? And you promised that meant you could get all the others.
 
  • #15
thanks buddy

your awesome
 
  • #16
It's "you're awesome". I'm correcting grammar tonight. Thanks.
 

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