How can the volume of a solid be found by rotating a region around a given line?

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SUMMARY

The volume of the solid formed by rotating the region bounded by the curves y = √[4]{x} and y = x around the line y = 1 can be calculated using the method of cylindrical shells or circular washers. The formula π∫r²h is applicable for calculating volumes, but it requires proper identification of the radius (r) and height (h) based on the specific setup of the problem. Sketching the region and the resulting solid is essential for correctly setting up the integral. Misapplication of the formula can lead to negative areas, indicating a misunderstanding of the geometric configuration.

PREREQUISITES
  • Understanding of integral calculus and volume calculation techniques.
  • Familiarity with the curves y = √[4]{x} and y = x.
  • Knowledge of the methods of cylindrical shells and circular washers.
  • Ability to sketch geometric regions and solids of revolution.
NEXT STEPS
  • Study the method of cylindrical shells for volume calculation.
  • Learn about the circular washer method for solids of revolution.
  • Practice sketching regions and solids generated by rotation around lines.
  • Explore integral setup techniques for volume problems involving curves.
USEFUL FOR

Students and educators in calculus, particularly those focusing on volume calculations of solids of revolution, as well as anyone seeking to deepen their understanding of integral applications in geometry.

regnar
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Find the volume of the solid obtained by rotating the region bounded by the given curves about the line y=1.

y = [tex]\sqrt[4]{x}[/tex] , y = xI couldn't figure out if a should subtract one from x or from y = [tex]\sqrt[4]{x}[/tex]. I don't know if I'm doing this right I tried subtracting it from x and got a negative area.
I also used this formula:
[tex]\pi[/tex][tex]\int[/tex]r^2 h
 
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regnar said:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the line y=1.

y = [tex]\sqrt[4]{x}[/tex] , y = x


I couldn't figure out if a should subtract one from x or from y = [tex]\sqrt[4]{x}[/tex]. I don't know if I'm doing this right I tried subtracting it from x and got a negative area.



I also used this formula:
[tex]\pi[/tex][tex]\int[/tex]r^2 h
This formula is to be used when your typical volume element is a circular disk of radius r and thickness h. It is not at all applicable in this problem. Have you drawn a sketch of the region bounded by the two curves? Have you drawn a sketch of the solid generated when the region is rotated around the line y = 1? These sketches are necessary in helping you understand how to set up your integral. In this problem there are two approaches: cylindrical shells or circular washers.
 

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