Multiple Integrals: Find Volume Bounded by Cylinders and Planes

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SUMMARY

The discussion focuses on calculating the volume bounded by the cylinders defined by the equation x² + y² = 1 and the planes y = z, x = 0, and z = 0 within the first octant. The correct setup for the triple integral is established as ∫ from x=0 to 1 ∫ from y=0 to √(1-x²) ∫ from z=0 to y dz dy dx. The integration process clarifies that the volume is determined by integrating y with respect to y first, leading to the conclusion that the initial assumption of using √(1-x²) directly in the double integral was incorrect.

PREREQUISITES
  • Understanding of multiple integrals in calculus
  • Familiarity with cylindrical coordinates
  • Knowledge of the first octant in three-dimensional space
  • Ability to perform integration with respect to multiple variables
NEXT STEPS
  • Study the application of triple integrals in cylindrical coordinates
  • Learn about volume calculations using multiple integrals
  • Explore the concept of bounded regions in three-dimensional calculus
  • Practice solving similar problems involving integration in the first octant
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Students and educators in calculus, particularly those focusing on multiple integrals and volume calculations in three-dimensional geometry.

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Homework Statement


Hello, I was wondering if someone could help me with the following. Supposed I am asked to find the volume bounded by the cylinders x^2+y^2=1 and the planes y = z, x = 0, z = 0 in the first octant.


Homework Equations


So this is what I tried to do. The boundaries should be: x is between 0 and 1 and y is between the squareroot of (1-x^2) and 0, or you can have y is between 0 and 1 and x is between the squareroot of (1-y^2) and 0. So wouldn't the double integral be the integral of

the squareroot of 1-x^2dydx, where you first evaluate it from 0 to the squareroot of (1-x^2), and then you evaluate it again from 0 to 1? Thanks!


The Attempt at a Solution

 
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Since you are in the first octant, yes, x runs between 0 and 1. For each x, then y runs from 0 up to the circle, [itex]y= \sqrt{1- x^2}[/itex]. Finally, for each x and y, z runs from 0 up to the plane z= y. The volume is given by
[tex]\int_{x=0}^1\int_{y=0}^{\sqrt{1-x^2}}\int_{z=0}^y dzdydx= \int_{x=0}^1\int_{y=0}^{\sqrt{1-x^2}}y dydx[/itex]<br /> No, that is NOT [itex]\sqrt{1- x^2}dydx[/itex]! You don't get the square root until after integrating with respect to y- and then, since the integral of ydy will involve y<sup>2</sup>, you don't really have a square root to integrate with respect to x![/tex]
 
Oh I see now! Thanks!
 

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