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Llana [10]
3 years ago
13

CALCULUS: Which of the following represents the volume of the solid formed by revolving the region bounded by the graphs of y =

img src="https://tex.z-dn.net/?f=x%5E%7B3%7D" id="TexFormula1" title="x^{3}" alt="x^{3}" align="absmiddle" class="latex-formula">, y = 1, and x = 3, about the line x = 3?
A. π * [27,1]∫ (3-∛y)² dy
B. π * ∛\int\limits^3_1 {(3-\sqrt[3]{y})^{2} } \, dy
C. None of these
Mathematics
2 answers:
Mademuasel [1]3 years ago
6 0

Answer:

π * [27,1]∫ (3-∛y)² dy

Step-by-step explanation:

Assoli18 [71]3 years ago
4 0

Disk method it is, since both given options integrate with respect to <em>y</em>.

First find where the boundaries of the region intersect.

• <em>y</em> = <em>x</em> ³ and <em>y</em> = 1 intersect at (1, 1)

• <em>y</em> = <em>x</em> ³ and <em>x</em> = 3 intersect at (3, 27)

• <em>y</em> = 1 and <em>x</em> = 3 intersect at (1, 3)

So the bounded region is the set of points

{ (<em>x</em>, <em>y</em>) | 1 ≤ <em>x</em> ≤ 3 and 1 ≤ <em>y</em> ≤ <em>x</em> ³ }

But since we're integrating with respect to <em>y</em>, rewrite this set so that <em>y</em> has constant limits:

{ (<em>x</em>, <em>y</em>) | ∛<em>y</em> ≤ <em>x</em> ≤ 3 and 1 ≤ <em>y</em> ≤ 27 }

Pick some point <em>y</em> in the interval [1, 27] and construct a disk centered at the given axis of revolution (<em>x</em> = 3). Such a disk will have radius 3 - ∛<em>y </em>because <em>x</em> = ∛<em>y</em> is the horizontal distance from the <em>y</em>-axis to the curve <em>y</em> = <em>x</em> ³ and <em>x</em> = 3 is itself 3 units away from the <em>y</em>-axis. Its height will be some small change in <em>y</em>, call it ∆<em>y</em>. Then the volume of this disk is

<em>π</em> (3 - ∛<em>y</em> )² ∆<em>y</em>

Now do the same thing for every <em>y</em> in [1, 27] - infinitely many of them! - and make ∆<em>y</em> very small, such that ∆<em>y</em> → d<em>y</em>. The volume of the solid is the sum total of the volumes of these infinitely many disks, given by the integral

\displaystyle\pi\int_1^{27}\left(3-\sqrt[3]{y}\right)^2\,\mathrm dy

and so the answer is A (assuming the limits of integration are listed from upper to lower).

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