## 5(a)

$$y^2 \, dx = \frac{1}{2}\int_0^3 (9-x^2)^2 \, dx = \frac{1}{2}\int_0^3 (81-18x^2+x^4) \, dx$$ | M1 | 2.1

$$= \frac{1}{2}\left[81x - 6x^3 + \frac{x^5}{5}\right]_0^3$$ | DM1 | 1.1b

$$= \frac{1}{2}\left(81 \cdot 3 - 6 \cdot 27 + \frac{243}{5}\right) = \frac{324}{5}$$ | A1 | 1.1b

$$\text{distance} = \frac{64.8}{18} = 3.6 \text{ m}$$ | A1* | 2.2a

(4)

### 5(a) Alternative

$$x y \, dy = \int_0^9 y\sqrt{9-y} \, dy$$ | M1 | 2.1

$$= \left[-\frac{2}{3}y(9-y)^{3/2} - \frac{4}{15}(9-y)^{5/2}\right]_0^9$$ | DM1 | 1.1b

$$= \frac{4}{15} \cdot 9^{5/2} = \frac{324}{5}$$ | A1 | 1.1b

$$\text{distance} = \frac{64.8}{18} = 3.6 \text{ m}$$ | A1* | 2.2a

(4)

**Notes:**

M1 | Use of $\frac{1}{2}\int y^2 \, dx$ or $\int xy \, dy$. Ignore any limits.

DM1 | Integrate and use correct limits (0 and 3 for $x$, 0 and 9 for $y$). Usual rules for integration: powers increasing by 1

A1 | Correct unsimplified expression.

A1* | Obtain given answer from correct working.

## 5(b)

Moments about O: | M1 | 3.1a

$$3.6W = 9W\bar{x}$$ | A1 | 1.1b

$$\bar{x} = 0.4$$ | A1 | 1.1b

(3)

**Notes:**

M1 | Complete method to obtain $\bar{x}$, e.g. by taking moments about O or by resolving and taking moments about a different point. Allow for an equation in $T$ or $\bar{x}$.

A1 | Correct unsimplified equation in $\bar{x}$

A1 | Correct only.