How to know the limits when finding the area under a parametric curve?

Question

In the last question on this paper (http://qualifications.pearson.com/content/dam/pdf/Advanced%20Extension%20Award/Mathematics/2013/Exam%20materials/Question-paper-Paper-1-June-2014.pdf), part (c), I integrated the correct function, but I put the limits as pi being the upper limit and zero being the lower limit. The markscheme (http://qualifications.pearson.com/content/dam/pdf/Advanced%20Extension%20Award/Mathematics/2013/Exam%20materials/Mark-scheme-Paper-1-June-2014.pdf), however, puts the limits as pi being the lower limit and 0 being the upper limit.

Answer 1

Now that I've actually read the problem . . .

Let $D_1$ be the region in the first quadrant, bounded by $C$.

The goal in part $(c)$ of the given problem is to find the area of $D_1$.

Let $C_1$ denote the part of the curve $C$ in the first quadrant.

Using the angle $\theta$ (from the diagram) as a parameter, $C_1$ can be parametrized by \begin{align*} x &= \sin(\theta) + (\pi-\theta)\cos(\theta)\\[4pt] y&= 1-\cos(\theta) + (\pi-\theta)\sin(\theta)\\[4pt] \end{align*} Since on the curve $C_1$

• $x$ varies from $0$ to $\pi$.
• $y$ is always nonnegative, varying continuously, as a function of $x$.

it follows that the area of $D_1$ can be expressed as $$ \int_{0}^\pi y\,dx \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $$ where $y$ is the value of the $y$-coordinate on the curve $C_1$ for a given value of $x$.

Switching to parametric form, we get \begin{align*} \int_{0}^{\pi} y\,dx &=\int_{\pi}^0 y\,\frac{dx}{d\theta}\,d\theta\\[2pt] &\qquad\;\;\;\bigl(\text{since}\;x=0\;\,\text{corresponds to}\;\,\theta=\pi\\[0pt] &\qquad\;\;\;\;\,\text{and}\;x=\pi\;\,\text{corresponds to}\;\,\theta=0\bigr)\\[2pt] &=\int_{\pi}^0 \bigl( 1-\cos(\theta) + (\pi-\theta)\sin(\theta) \bigr) \bigl(-(\pi-\theta)\sin(\theta)\bigr)\,d\theta\\[6pt] &=-\int_{\pi}^0 \bigl( 1-\cos(\theta) + (\pi-\theta)\sin(\theta) \bigr) \bigl((\pi-\theta)\sin(\theta)\bigr)\,d\theta\\[6pt] &=\int_0^\pi \bigl( 1-\cos(\theta) + (\pi-\theta)\sin(\theta) \bigr) \bigl((\pi-\theta)\sin(\theta)\bigr)\,d\theta\\[6pt] &\qquad\;\;\;\;\;\vdots\\[6pt] &=\frac{\pi(\pi^2+\,3)}{6}\\[6pt] \end{align*}

To recap, as to the question of why the integral initially has $\theta$ going from $\pi$ to $0$, it's simply to match the corresponding values of $x$ which go from $x=0$ to $x=\pi$.