Volume of a solid w/known cross section

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SUMMARY

The discussion focuses on calculating the volume of a solid with a base region bounded by the curves y=x, y=(x-1)^2, and the line x=1, using semicircular cross sections perpendicular to the x-axis. The user correctly identifies the volume formula for a semicircular disk as (1/2)π(r^2)(h) and determines the intersection points of the curves to establish the limits for integration. The Riemann Sum is formulated as Σ (π/4)(-x² + 3x - 3)²Δx, leading to the definite integral (π/4) ∫[0.382, 1] (-x² + 3x - 3)² dx. The lower limit of integration is confirmed as (3 - √5)/2.

PREREQUISITES
  • Understanding of Riemann sums and definite integrals
  • Knowledge of semicircular cross-sectional volume calculations
  • Familiarity with curve intersection and bounding regions
  • Basic calculus concepts, including limits and integration techniques
NEXT STEPS
  • Study the application of Riemann sums in volume calculations
  • Learn about semicircular cross sections and their properties
  • Explore methods for finding intersections of curves analytically
  • Review definite integral evaluation techniques in calculus
USEFUL FOR

Students in calculus courses, educators teaching volume calculations, and anyone interested in understanding the application of integrals in geometric contexts.

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Homework Statement



The base region of a solid is bounded by y=x, y=(x-1)^2, and x = 1.

The cross sections are semicircles perpendicular to the x-axis.

Write a riemann sum and definite integral.

Homework Equations





The Attempt at a Solution



First, I wrote down the formula for a semicircular disk's volume. 1/2(pi(r^2)(h))

I then found the intersection of y=x and y=(x-1)^2 to be .382 and another value that was greater than 1, so I ditched it.

I then wrote down the diameter of any given disk as x - (x-1)^2 or -x^2 + 3x - 3, so radius is half of that, and I defined the height of each disk to be delta x.

So, I wrote the Riemann Sum as: (limit as delta x approaches 0)

\Sigma \frac{\pi}{4}(-x^{2}+3x-3)^{2}\Delta x

And therefore wrote a definite integral as:

\frac{\pi}{4} \int^{1}_{.382} (-x^{2}+3x-3)^{2} dx

Did I do this right?
 
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It looks good. The true value for the lower limit of integration is, \displaystyle \frac{3-\sqrt{5}}{2}\,.
 

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