Standard +0.3 This is a straightforward volume of revolution question requiring integration with respect to y. Students must apply the standard formula V = π∫x²dy, substitute x = 8/y² - 2, expand (8/y² - 2)², and integrate term-by-term using standard power rules. While it requires careful algebraic manipulation and finding limits from the diagram, it follows a routine procedure with no conceptual surprises, making it slightly easier than average.
\includegraphics{figure_5}
The diagram shows part of the curve \(x = \frac{8}{y^2} - 2\), crossing the \(y\)-axis at the point \(A\). The point \(B (6, 1)\) lies on the curve. The shaded region is bounded by the curve, the \(y\)-axis and the line \(y = 1\). Find the exact volume obtained when this shaded region is rotated through \(360°\) about the \(y\)-axis. [6]
Integral of \(x^2 = \frac{64y^{-3}}{-3} - \frac{32y^{-1}}{-1} + 4y\)
B1, B1, B1
All co
Uses limits 1 to 2; \(\to 67\frac{2}{7}\pi\)
M1, A1
Uses 1 to 2 or 2 to 1; co
$x = \frac{8}{y^2} - 2$; at $y = 0, y = 2$
$\to x^2 = \frac{64}{y^4} - \frac{32}{y^2} + 4$ | B1 | co
Integral of $x^2 = \frac{64y^{-3}}{-3} - \frac{32y^{-1}}{-1} + 4y$ | B1, B1, B1 | All co
Uses limits 1 to 2; $\to 67\frac{2}{7}\pi$ | M1, A1 | Uses 1 to 2 or 2 to 1; co
\includegraphics{figure_5}
The diagram shows part of the curve $x = \frac{8}{y^2} - 2$, crossing the $y$-axis at the point $A$. The point $B (6, 1)$ lies on the curve. The shaded region is bounded by the curve, the $y$-axis and the line $y = 1$. Find the exact volume obtained when this shaded region is rotated through $360°$ about the $y$-axis. [6]
\hfill \mbox{\textit{CAIE P1 2012 Q5 [6]}}