Volume of Solid of Revolution - Past HSC Exam Question (1 Viewer)

phoenix159 · Nov 3, 2013

A bowl is formed by rotating the part of the curve y = x^4 / 4 between x=0 and x=2 about the y axis.
Find the volume of the bowl.

From: 2002 Mathematics Exam (2 Unit)

Shazer2 · Nov 4, 2013

$x = \sqrt [4]{4y} \\ \\ \indent x^{2} = (4y)^{\frac {1}{8}} \\ \\ \indent V = \pi \int_{0}^{2} (4y)^{\frac {1}{8}} $ dy $ \\ \\ \indent V = \pi \left[\frac {8(4y)^{\frac {9}{8}}}{9}}\right]_0^2 \\ \\ \indent V = \frac {8\pi \sqrt [8]{8^{9}}}{9} $ units$^{3}$

Pretty sure that might be wrong, but gave it a go.

bottleofyarn · Nov 4, 2013

I found the paper and there's a diagram which clears some ambiguity here (2002 q6c).
This means that you should integrate with respect to y from 0 to 4 since you're rotating about the y axis.
Then, you want the volume between the curve and y=4:
$V=\int_0^4{\pi x^2}\,dy=\pi \int_0^4{2\sqrt{y}}\,dy$
which is simple enough and gives you 32pi/3.

Had a glance at the marking report and the method should be right.

Volume of Solid of Revolution - Past HSC Exam Question (1 Viewer)

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