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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.3.69b
Chapter 6, Problem 6.3.69b

A right circular cylinder with height R and radius R has a volume of VC=πR^3 (height = radius).


b. Find the volume of the hemisphere that is inscribed in the cylinder with the same base as the cylinder. Express the volume in terms of VC.

Verified step by step guidance
1
Identify the dimensions of the hemisphere inscribed in the cylinder. Since the hemisphere shares the same base as the cylinder, its radius is equal to the radius of the cylinder, which is \(R\).
Recall the formula for the volume of a hemisphere: \(V_H = \frac{2}{3} \pi r^3\). Substitute \(r = R\) to get \(V_H = \frac{2}{3} \pi R^3\).
Recall the volume of the cylinder given as \(V_C = \pi R^3\) (since height = radius = \(R\)).
Express the volume of the hemisphere \(V_H\) in terms of the cylinder volume \(V_C\) by dividing \(V_H\) by \(V_C\): \(\frac{V_H}{V_C} = \frac{\frac{2}{3} \pi R^3}{\pi R^3}\).
Simplify the expression to find \(V_H\) as a fraction of \(V_C\), which will give the volume of the hemisphere in terms of the cylinder volume.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Volume of a Cylinder

The volume of a right circular cylinder is calculated by multiplying the area of its base by its height. For a cylinder with radius R and height R, the volume is V = πR² × R = πR³. This formula provides the reference volume VC used in the problem.
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Finding Volume Using Disks

Volume of a Hemisphere

A hemisphere is half of a sphere. The volume of a sphere with radius r is (4/3)πr³, so the volume of a hemisphere is half of that, (2/3)πr³. Understanding this formula is essential to find the volume of the hemisphere inscribed in the cylinder.
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Inscribed Shapes and Radius Relationship

An inscribed hemisphere inside a cylinder with the same base means the hemisphere's radius equals the cylinder's radius R. Recognizing this relationship allows substitution of R into the hemisphere volume formula and expressing the result in terms of VC.
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Related Practice
Textbook Question

Use the region R that is bounded by the graphs of y=1+√x,x=4, and y=1 complete the exercises.


Region R is revolved about the x-axis to form a solid of revolution whose cross sections are washers.


b. What is the inner radius of a cross section of the solid at a point x in [0, 4]?

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Textbook Question

Equal integrals Without evaluating integrals, explain the following equalities. (Hint: Draw pictures.)


b. ∫²₀(25−(x²+1)²) dx = 2∫₁⁵ y√y−1 dy

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Textbook Question

Different axes of revolution Suppose R is the region bounded by y=f(x) and y=g(x) on the interval [a, b], where f(x)≥g(x).


b. How is this formula changed if x0>b?

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Textbook Question

Deceleration A car slows down with an acceleration of a(t) = −15 ft/s². Assume v(0)=60 ft/s,s(0)=0, and t is measured in seconds.


b. How far does the car travel in the time it takes to come to rest?

Textbook Question

Volume of a sphere Let R be the region bounded by the upper half of the circle x²+y² = r² and the x-axis. A sphere of radius r is obtained by revolving R about the x-axis.


b. Repeat part (a) using the disk method.

Textbook Question

Determine whether the following statements are true and give an explanation or counterexample. 


b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.