5.17 Areas with the y-axis and volumes of revolutionIB Maths: Analysis and Approaches HL: Revision notes

For the region between a curve and the $y$-axis, swap the roles of $x$ and $y$: make $x$ the subject, $x=g(y)$, and integrate with respect to $y$ between horizontal lines $y=a$ and $y=b$: $A=\int_a^b |x|\,dy=\int_a^b |g(y)|\,dy.$ Example: for $y=\ln x$ between $y=0$ and $y=2$, $x=e^{y}$ and $A=\int_0^2 e^{y}\,dy=e^{2}-1$.

The limits are $y$-values. If the curve is to the left of the $y$-axis ($x<0$), the integral is negative, so take the modulus, exactly as for regions below the $x$-axis.

Sometimes a region is simpler in horizontal strips. Write both boundaries as $x=g_1(y)$ (right) and $x=g_2(y)$ (left); then $A=\int_c^d \left(g_1(y)-g_2(y)\right)dy.$ Example: between $y=\sqrt{x}$ ($x=y^{2}$) and $y=\frac{x}{2}$ ($x=2y$), meeting at $y=0$ and $y=2$: $A=\int_0^2(2y-y^{2})\,dy=\frac43$, the same as $\int_0^4\left(\sqrt{x}-\frac{x}{2}\right)dx$.

Rotating $y=f(x)$, $a\le x\le b$, through $2\pi$ about the $x$-axis makes a solid whose cross-sections are discs of radius $y$: $V=\pi\int_a^b y^{2}\,dx.$ Example: $y=\frac{2}{\sqrt{x}}$ from $x=1$ to $x=4$ gives $V=\pi\int_1^4\frac{4}{x}\,dx=4\pi\ln4=8\pi\ln2$.

This formula is in the formula booklet; what you must do is square $y$ correctly and choose the right limits.

Rotating about the $y$-axis, the discs have radius $x$, so make $x$ the subject and use $V=\pi\int_c^d x^{2}\,dy.$ Often you only need $x^{2}$, so there is no need to take a square root: for $y=\frac{x^{2}}{4}$, $x^{2}=4y$ directly.

This is the natural model for bowls, vases and containers standing on a table: the depth of liquid $h$ becomes the upper limit, giving the volume as a function of depth (for the bowl $y=\frac{x^{2}}{4}$, $V=2\pi h^{2}$).

When the region between two curves is rotated, the solid has a hole. Each slice is a washer: outer radius $y_1$, inner radius $y_2$: $V=\pi\int_a^b\left(y_1^{2}-y_2^{2}\right)dx\quad\text{or}\quad V=\pi\int_c^d\left(x_1^{2}-x_2^{2}\right)dy.$ Steps: find where the curves meet (these are the limits), decide which curve is further from the axis of rotation, then square each separately.

For the region between $y=\sqrt{x}$ and $y=\frac{x}{2}$: about the $x$-axis $V=\pi\int_0^4\left(x-\frac{x^{2}}{4}\right)dx=\frac{8\pi}{3}$; about the $y$-axis $V=\pi\int_0^2(4y^{2}-y^{4})\,dy=\frac{64\pi}{15}$.