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Problem 65 Hard Difficulty
(a) Cavalieri's Principle states that if a family of parallel planes gives equal cross-section areas for two solids $ S_1 $ and $ S_2 $ then the volumes of $ S_1 $ and $ S_2 $ are equal. Prove this principle.
(b) Use Cavalieri's Principle to find the volume of the oblique cylinder shown in the figure.
Answer
(a) Volume $\left(S_{1}\right)=\int_{0}^{h} A(z) d z=$ Volume $\left(S_{2}\right)$ since the cross-sectional area $A(z)$ at height $z$ is the same for both solids.
(b) By Cavalieri's Principle, the volume of the cylinder in the figure is the same as that of a right circular cylinder with radius $r$ and height $h,$ that is, $\pi r^{2} h$
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Topics
Applications of Integration
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Video Transcript
So for this problem, we're gonna let A s one and A s to represent the areas of the cross sections of the solids s one and s two. So the volume of s one will be equal to are bound the integral of our bounds a one to be one of a s one DX and similarly, we have a volume of the S two, which is a 22 b two a s, two d x some. Based on this, we're assuming that s one equals s two. Because the planes of the cross sections are parallel, we know that the bounds of the integral czar equal so a one equals a two and B one equals b two. So based on this, since these two areas are equal to each other and then we also have that the bounds are equal to each other. We can conclude that a one B one A s one D x is equal to a to B two a s, two DX Um So what that ultimately means is that V S one is equal to V S two than for Part B. Um, we have a given cylinder C one And then we have another cylinder, C two with height each and radius arm, and we can assume that they're lower bases are in the same plane. Um, using the principle that we just discussed. What we have is that the volume of the first one equals the Honourable from zero to h of a one x dx, and that's going to be equal r squared pi times in a row from zero to h dx. And that's just r squared h Then we also do is V two, and we end up getting that. This is going to give us the same result of R squared pi H, and we already knew that their bounds were equal. So we see that since their volumes are equal, um, we can set up those cylinders on the same plane, and it's clear that they have the same intersection with corresponding plane. Um, and we see that their volumes are equal, which is what we wanted to show. And this is R squared pi H, by the way,
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Topics
Applications of Integration
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Anna Marie V.
Campbell University
Samuel H.
University of Nottingham
Joseph L.
Boston College