AHL 5.17—Areas under curve onto y-axis, volume of revolution (about x and y axes)

The region is bounded by the curve $y=\sqrt{x}$ and the y-axis, so we integrate horizontal strips with respect to $y$. We need to express $x$ as a function of $y$.

From $y=\sqrt{x}$, squaring both sides gives $x=y^2$. So $x(y)=y^2$.

The region is bounded between $y=0$ and $y=3$. Since we integrate with respect to $y$, the limits are $y$-values.

The area is $A={\int }_0^{3}x(y)dy={\int }_0^3{y}^{2}dy$. This matches the first option.

We need an integral of the form ${\int }_a^{b}x(y)dy$ with $x$ expressed as a function of $y$ and limits in terms of $y$.

The option ${\int }_0^3\sqrt{y}dy$ is incorrect because $x(y)=y^2$, not $\sqrt{y}$. Substituting $\sqrt{y}$ confuses the original function with the rearranged one.

Both ${\int }_0^9\sqrt{x}dx$ and ${\int }_0^9{y}^{2}dy$ use $x$-values as limits (since $x$ ranges from $0$ to $9$ on this curve) but the region is defined by $y$-boundaries $0$ to $3$, not $0$ to $9$.

${\int }_0^3{y}^{2}dy$ correctly uses $x(y)=y^2$ as the integrand and $y$-limits from $0$ to $3$.

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

$[f(x){]}^2={\left(\frac{\ln x}{x^2}\right)}^2=\frac{(\ln x{)}^2}{x^4}$.

The limits are $x=1$ to $x=e$ as given. The integral is $V=\pi {\int }_1^e\frac{(\ln x{)}^2}{x^4}dx$.

This matches the first option. The $\pi$ factor is kept outside, and the integrand is $[f(x){]}^2$, not $f(x)$.

The volume formula requires squaring $f(x)$ and multiplying by $\pi$. Any option missing the square or using the wrong constant is wrong.

$V=\pi {\int }_1^e\frac{\ln x}{x^2}dx$ uses $f(x)$ rather than $[f(x){]}^2$. The disk radius must be squared.

$V=\pi {\int }_1^e\frac{(\ln x{)}^2}{x^2}dx$ squares the numerator but not the denominator. Correctly, $(x^2{)}^2=x^4$, not $x^2$.

$V=2\pi {\int }_1^e\frac{(\ln x{)}^2}{x^4}dx$ uses $2\pi$ instead of $\pi$. The factor $2\pi$ arises in the shell method, not the disk method.

$V=\pi {\int }_1^e\frac{(\ln x{)}^2}{x^4}dx$ correctly applies the disk method formula.