Finding the Volume of a Solid In Exercises 1-6,set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis. y = -x + 1
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Textbook Solutions for Calculus: Early Transcendental Functions
Question
HOW DO YOU SEE IT? Use the graph to match the integral for the volume with the axis of rotation.
(a) \(V=\pi \int_{0}^{b}\left(a^{2}-[f(y)]^{2}\right) d y\) (i) x-axis
(b) \(V=\pi \int_{0}^{a}\left(b^{2}-[b-f(x)]^{2}\right) d x\) (ii) y-axis
(c) \(V=\pi \int_{0}^{a}[f(x)]^{2} d x\) (iii) x = a
(d) \(V=\pi \int_{0}^{b}[a-f(y)]^{2} d y\) (iv) y =b
Text Transcription:
V=pi int_0^b (a^2 - [f(y)]^2) dy
V=pi int_0^a (b^2-[b-f(x)]^2) dx
V=pi int_0^a [f(x)]^2 dx
V=pi int_0^b [a-f(y)]^2 dy
Solution
The first step in solving 7.2 problem number 54 trying to solve the problem we have to refer to the textbook question: HOW DO YOU SEE IT? Use the graph to match the integral for the volume with the axis of rotation. (a) \(V=\pi \int_{0}^{b}\left(a^{2}-[f(y)]^{2}\right) d y\) (i) x-axis(b) \(V=\pi \int_{0}^{a}\left(b^{2}-[b-f(x)]^{2}\right) d x\) (ii) y-axis(c) \(V=\pi \int_{0}^{a}[f(x)]^{2} d x\) (iii) x = a(d) \(V=\pi \int_{0}^{b}[a-f(y)]^{2} d y\) (iv) y =bText Transcription:V=pi int_0^b (a^2 - [f(y)]^2) dyV=pi int_0^a (b^2-[b-f(x)]^2) dxV=pi int_0^a [f(x)]^2 dxV=pi int_0^b [a-f(y)]^2 dy
From the textbook chapter Volume:The Disk Method you will find a few key concepts needed to solve this.
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