# Ask Uncle Colin: A Parametric Integration

Dear Uncle Colin,

I have the parametric equations $x = (t+1)^2$ and $y = \frac{1}{2}t^3 + 3$ and the lines $y = 16 - x$ and $x=1$. I need to find the area enclosed by the curve, these two lines and the $x$-axis, but my answer doesn’t agree with the book’s! Can you help?

- Frankly I’m Surprised How Errors Result

Hi, FISHER, and thanks for your message!

Before we begin, let’s sketch the diagram:

There are clearly two bits to integrate: a wiggly bit on the left and a triangle on the right. The triangle is probably easiest to begin with.

## The triangle

We need to find the intersection point of the curve and the line, so that $(t+1)^2 = 13 - \frac{1}{2}t^3$.

Doubling and expanding: $2t^2 + 4t + 2 = 26 - t^3$, or $t^3 + 2t^2 + 4t - 24 = 0$.

This factorises as $(t-2)(t^2 + 4t + 12)$, and the only real solution to that is at $t=2$.

That corresponds to the point $(9,7)$; we can immediately see that the triangle has height of 7 and base of 7, and therefore an area of $\frac{49}{2}$.

## The wiggle

I can see two reasonable ways to solve this, so let’s try both. The first, and I think easiest, is parametric integration; the second is to make it cartesian and integrate directly.

### Parametric

When $x=1$, $t=0$, so we would need to calculate $\int_0^2 y \diff{x}{t} \dt$.

This works out to be $\int_0^2 (t^3 + 6)(t+1) \dt$, or $\int_0^2 t^4 + t^3 + 6t + 6 \dt$.

That’s $\left[ \frac{1}{5}t^5 + \frac{1}{4}t^4 + 3t^2 + 6t\right]_0^2$, which evaluates to $\frac{32}{5} + 4 + 12 + 12$; I get $\frac{172}{5}$.

### Cartesian

Alternatively, we can rearrange the first equation to find that $t = x^{\frac{1}{2}}-1$.

In this case, $y = \frac{1}{2} \left( x^{\frac{1}{2}}-1\right)+3$, which we can expand: $y = \frac{1}{2} \left( x^{\frac{3}{2}} - 3x + 3x^{\frac{1}{2}} + 5\right)$, bringing the 3 inside the bracket.

We need $\int_1^9 y \dx$, which is $\frac{1}{2}\left[ \frac{2}{5}x^{\frac{5}{2} } - \frac{3}{2}x^2 + 2x^\frac{3}{2} + 5x\right]_1^9$.

Evaluating this, I get $\frac{1}{2}\left[ \frac{2}{5}\times 242 - \frac{3}{2}\times 80 + 2 \times 26 + 5 \times 8 \right]$.

This gives $\frac{242}{5} - 60 + 26 + 20$, or $\frac{172}{5}$ again. Phew.

## Altogether

Finally, we need to add $\frac{172}{5}$ to $\frac{49}{2}$, which gives $\frac{344 + 245}{10} = 58.9$ square units.

Hope that helps!

- Uncle Colin

## A selection of other posts

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