AHL 5.12—Areas under a curve onto x or y axis. Volumes of revolution about x and y

The area between $x=y^2$ and the y-axis is $A={\int }_1^{3}x(y)dy$ since $x(y)=y^2\ge 0$ on $[1,3]$.

$A={\int }_1^3{y}^{2}dy$.

$A={\left[\frac{y^3}{3}\right]}_1^3=\frac{27}{3}-\frac{1}{3}=\frac{26}{3}$.

The area is $\frac{26}{3}$ square units.

$\frac{8}{3}$ is the area from $y=0$ to $y=2$, using wrong limits.

$\frac{80}{3}$ results from mistakenly using x-limits $x=1$ to $x=3$ instead of y-limits.

$8$ comes from evaluating ${\left[y^3\right]}_1^3$ without dividing by 3, forgetting the coefficient.

Only careful integration with correct y-limits gives $\frac{26}{3}$.

For rotation about the x-axis, $V=\pi {\int }_a^b[f(x){]}^{2}dx$.

Here $f(x)=\sqrt{x}$, so $[f(x){]}^2=x$.

$V=\pi {\int }_0^{4}xdx$.

This matches option A exactly.

$\pi {\int }_0^4\sqrt{x}dx$ forgets to square $f(x)$ before integrating.

$\pi {\int }_0^4{x}^{2}dx$ squares $x$ instead of $f(x)=\sqrt{x}$, an incorrect substitution.

${\int }_0^{4}xdx$ omits the factor of $\pi$ required for a circular cross-section.

Only squaring $\sqrt{x}$ correctly and including $\pi$ gives option A.