2. Locate the centroid of the given volumes.x1 ft-3 ft06 PROB 31(a)-(x-1)xm06 PRO 34y²-0.52(b) [Submit]2my

The equation of the cone is z=y6y-1 Now consider the elementary strip dV=πz2dy =πy6y-12dy…

Answered: 2. Locate the centroid of the given…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

2. Locate the centroid of the given volumes.
x
1 ft
-3 ft
06 PROB 31
(a)
-(x-1)
x
m
06 PRO 34
y²-0.52
(b) [Submit]
2m
y

Expert Solution

Step 1: Finding the centroid of cone

The equation of the cone is $z = \frac{y}{6} (y - 1)$

Now consider the elementary strip

$d V = π z^{2} d y = π {(\frac{y}{6} (y - 1))}^{2} d y$

Therefore the centroid

$y = \frac{\int_{0}^{3} y d V}{\int_{0}^{3} d V} = \frac{\int_{0}^{3} y π {(\frac{y}{6} (y - 1))}^{2} d y}{\int_{0}^{3} π {(\frac{y}{6} (y - 1))}^{2} d y} = \frac{\int_{0}^{3} y \frac{{(y^{2} - y)}^{2}}{36} d y}{\int_{0}^{3} \frac{{(y^{2} - y)}^{2}}{36} d y} = \frac{\int_{0}^{3} \frac{y (y^{4} - 2 y^{3} + y^{2})}{36} d y}{\int_{0}^{3} \frac{(y^{4} - 2 y^{3} + y^{2})}{36} d y} = \frac{\int_{0}^{3} y^{5} - 2 y^{4} + y^{3} d y}{\int_{0}^{3} y^{4} - 2 y^{3} + y^{2} d y} = \frac{{[\frac{y^{6}}{6} - \frac{2 y^{5}}{5} + \frac{y^{4}}{4}]}_{0}^{3}}{{[\frac{y^{5}}{5} - \frac{y^{4}}{2} + \frac{y^{3}}{3}]}_{0}^{3}} = \frac{121.5 - 97.5 + 20.25}{48.6 - 40.5 + 9} = \frac{44.55}{17.1} = 2.6052 ∴ y = 2.6052$