The equation of a curve is \(y = \frac{9}{2-x}\).
Find the volume obtained when the region bounded by the curve, the coordinate axes and the line \(x = 1\) is rotated through 360° about the x-axis.
Solution
The volume \(V\) of the solid of revolution is given by the formula:
\(V = \pi \int_0^1 y^2 \, dx\)
Substitute \(y = \frac{9}{2-x}\) into the formula:
\(V = \pi \int_0^1 \left(\frac{9}{2-x}\right)^2 \, dx\)
\(V = \pi \int_0^1 \frac{81}{(2-x)^2} \, dx\)
Integrate \(\frac{81}{(2-x)^2}\):
\(\int \frac{81}{(2-x)^2} \, dx = -81 \left(2-x\right)^{-1} \times (-1)\)
\(= 81 \left(2-x\right)^{-1}\)
Evaluate from 0 to 1:
\(V = \pi \left[ 81 \left(2-1\right)^{-1} - 81 \left(2-0\right)^{-1} \right]\)
\(= \pi \left[ 81 \times 1 - 81 \times \frac{1}{2} \right]\)
\(= \pi \left[ 81 - 40.5 \right]\)
\(= \pi \times 40.5\)
\(= \frac{81\pi}{2}\)
Log in to record attempts.