# Question 5:

## Part 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $x$-axis: $(\rho)\int x \times 2\sqrt{r^2-x^2}\,dx$ OR Using $y$-axis: $(\rho)\frac{1}{2}\int 2\left(\sqrt{r^2-x^2}\right)^2\,dx$ | M1 | Use of correct integral. Limits not needed here. Accept $x$-axis: $k\int x\sqrt{r^2-x^2}\,dx$; $y$-axis: $k\int r^2-x^2\,dx$ |
| $x$-axis: $=-\frac{2}{3}(\rho)(r^2-x^2)^{\frac{3}{2}}$ OR $y$-axis: $=(\rho)\left(xr^2-\frac{x^3}{3}\right)$ | A1 | Correct integration, ignore limits. Correct expression. |
| $=\frac{2}{3}(\rho)r^3$ | A1 | Correct use of limits, $0$ and $r$ or $-r$ and $r$ |
| Using $x$-axis: $\frac{1}{2}\pi r^2\rho\bar{x}=\rho\int_0^r 2xy\,dx$ OR Using $y$-axis: $\frac{1}{2}\pi r^2\rho\bar{y}=\rho\frac{1}{2}\int_{-r}^{r}y^2\,dx$ or $\frac{1}{2}\pi r^2\rho\bar{y}=\rho\int_0^r y^2\,dx$ | M1 | Complete method to obtain distance. Use of correct formula consistent with axis and limits used. $\rho$ must appear on both sides or neither. |
| $\bar{x}=\dfrac{\frac{2}{3}r^3}{\frac{1}{2}\pi r^2}=\dfrac{4r}{3\pi}$ * | A1* | Obtain given answer from correct working |

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## ALT 1 Part 5a (Parametric):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=r\cos\theta,\ y=r\sin\theta$; Using $x$-axis: $2r^3\int_0^{\frac{\pi}{2}}\sin^2\theta\cos\theta\,d\theta$ | M1 | Use of correct integral. Accept $kr^3\int\sin^2\theta\cos\theta\,d\theta$ |
| $=2r^3\left[\dfrac{\sin^3\theta}{3}\right]_0^{\frac{\pi}{2}}$ | A1 | Correct integration, ignore limits. |
| $=\frac{2}{3}r^3$ | A1 | Correct use of limits |
| $\bar{x}=\dfrac{\frac{2}{3}r^3}{\frac{1}{2}\pi r^2}$ | M1 | Complete method. $\rho$ must appear on both sides or neither. |
| $=\dfrac{4r}{3\pi}$ * | A1* | Obtain given answer from correct working |

**[5]**

## ALT 2 Part 5a (Using $y$-axis):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r^3\int_0^{\frac{\pi}{2}}\sin^3\theta\,d\theta$ | M1 | Accept $kr^3\int\sin^3\theta\,d\theta$ |
| $r^3\int_0^{\frac{\pi}{2}}(1-\cos^2\theta)\sin\theta\,d\theta = r^3\left[-\cos\theta+\dfrac{\cos^3\theta}{3}\right]_0^{\frac{\pi}{2}}$ | A1 | Correct integration, ignore limits. |
| $=\frac{2}{3}r^3$ | A1 | Correct use of limits |
| $\bar{x}=\dfrac{\frac{2}{3}r^3}{\frac{1}{2}\pi r^2}$ | M1 | Complete method. $\rho$ must appear on both sides or neither. |
| $=\dfrac{4r}{3\pi}$ * | A1* | Obtain given answer from correct working |

**[5]**

## Part 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: large $8\pi a^2$, small removed $2\pi a^2$, small added $2\pi a^2$; Distances from $AC$: large $\frac{16a}{3\pi}$, small removed $\frac{8a}{3\pi}$, small added $(-)\frac{8a}{3\pi}$ | B1, B1 | Correct mass ratios; Correct distances |
| Moments about $AC$: $8\pi a^2\times\dfrac{16a}{3\pi}-2\pi a^2\times\dfrac{8a}{3\pi}-2\pi a^2\times\dfrac{8a}{3\pi}=8\pi a^2 d$ | M1 | All terms required. Dimensionally correct or equivalent for parallel axis. Condone sign errors. |
| $\dfrac{96a}{3\pi}=8d \Rightarrow d=\dfrac{4a}{\pi}$ * | A1* | Obtain given value from correct working. Need to see at least some simplification. |

**[5]**

## Part 5c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about perpendicular axis through $A$: $4a\times 8\pi a^2 - 2a\times 2\pi a^2 + 6a\times 2\pi a^2 = 8\pi a^2\bar{x}$ | M1, A1ft, A1ft | Dimensionally correct, all terms. Unsimplified equation with at most one error. Correct unsimplified equation. Follow their mass ratio. |
| $\Rightarrow \bar{x}=5a$ | A1 | Correct only. If measured from $B$, distance is $a$ |
| Correct use of trig: $\tan\theta=\dfrac{d}{\bar{x}}$ or $\tan\theta=\dfrac{\bar{x}}{d}$ where $\bar{x}$ is distance from $A$ | M1 | $\tan\theta=\frac{d}{\bar{x}}$ or $\tan\theta=\frac{\bar{x}}{d}$ |
| $\tan\theta=\dfrac{4}{5\pi}$ | A1 | Only |

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