# 2011 STEP 2 Q8

The third part of the question asks you to integrate using the formula given to find the area between the curve, the x axis and limits of $\pi$ and $\frac{\pi}{2}$

The mark scheme uses $\frac{\pi}{2}$ as the upper limit while $\pi$ as the lower limit - how would you know to do this?

As I used the upper limit as $\pi$ the answers I obtained were the same as the mark scheme except negative.

I realise that area cannot be negative but why is it that using $\pi$ as the upper limit gives the negative answer? - the curve does not go below the x axis so we are not expecting a negative answer.

### Try looking at your sketch

If you look at your sketch (assuming it is correct), you should see that $t=\pi$ corresponds to $(1,0)$ and $t=\frac{\pi}{2}$ corresponds to $\left(\frac{\pi}{2},1\right)$, so if we want to integrate to find the area between the curve and the $x$-axis we want to integrate from $x=1$ to $x=\frac{\pi}{2}$ which explains the order of the limits.

### I found that $t= \pi$

I found that $t= \pi$ corresponds to $(-1, \pi)$. I realise that we want the greater x coordinate as the upper limit - in this case it is $x=\frac{\pi}{2}$ (corresponding to $t=\frac{\pi}{2}$ as the upper limit) however what does it actually mean to find the area in this case?
When I sketched the graph; it is not actually a function. There are values where the graph is one to many - between $x=1$ and $x=\frac{\pi}{2}$ hence finding the area does not make sense when the graph is not a function.

### Ah yes, I have misunderstood

Ah yes, I have misunderstood my solution for this question. You're quite right that $t=\pi$ corresponds to $(-1,\pi)$. However my point still stands in that you want the value of $t$ with the greater $x$-coordinate to be the upper limit of the integral (though you seem to have understood this).

If we consider the curve described by $B$ restricted to $\pi\geq t\geq\frac{\pi}{2}$, then this is a one-to-one (usually referred to as injective at university level) function so, using the integral to find the area between it and the $x$-axis does make sense.

### Actually, ignore my comment

Actually, ignore my comment about 'injective' as you can have (what you would call) 'one-to-one' functions that are not injective.

### I understand that the curve

I understand that the curve is one to one when $\pi\geq t\geq\frac{\pi}{2}$ - this is just finding the area between $x=-1$ and $x= \frac{\pi}{2}$. However in the next part of the question, you have to find the area between the curve and the x axis - why isn't this the same answer as the previous part?
Surely since $x=\frac{\pi}{2}$ is the maximum x value then integrating when $\pi\geq t\geq\frac{\pi}{2}$ or when $\pi\geq t\geq 0$ (curve not one to one in this range) must give the same answer? - in both cases you are integrating between $x=-1$ and $x= \frac{\pi}{2}$

### Let's denote the region, that

Let's denote the region, that the integral in the second to last part calculates the area of, as $S_{1}$ (i.e. the region enclosed by the curve when restricted to $\pi\geq t \geq \frac{\pi}{2}$, the $x$-axis, and the lines $x=-1,\frac{\pi}{2}$) and denote the area of this region by $\left|S_{1}\right|$ so $$S_{1}=\left\{(x,y)\in\mathbb{R}^2:-1\leq x\leq \cos t + t\sin t,\,0\leq y \leq \sin t +t\cos t \text{ where } \pi\geq t \geq \frac{\pi}{2} \right\},$$ and $$\left|S_1\right|=\int_{\pi}^{\frac{\pi}{2}}y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t.$$ Let's denote the region enclosed by the curve, the $x$-axis and the line $x=\frac{\pi}{2}$ by $S_2$ so $$S_2=\left\{(x,y)\in\mathbb{R}^2:\cos t+t\sin t\leq x\leq\frac{\pi}{2},\, 0\leq y \leq \sin t +t\cos t \text{ where } 0\leq t \leq \frac{\pi}{2} \right\}.$$ Finally, denote the region of the unit semi-circular disc by $S_3$ so $$S_3=\left\{(x,y)\in\mathbb{R}^2:y\geq 0,\,x^2+y^2\leq 1\right\}.$$I think, in the final part, they want you to calculate the area of the region swept out by the string which is given by $$|S_1|-|S_2|-|S_3|$$

### As I'm sure you're aware, we

As I'm sure you're aware, we define the integral over a negative range (i.e. $[a,b]$ where $b < a$) by $$\int^b_a f(x)\,\mathrm{d}x=-\int^a_b f(x)\,\mathrm{d}x,$$ so if you decided to integrate over $[\pi,0]$ you would actually get (thinking about how the $x$-coordinate varies with $t$) $$\int^0_{\pi}y\frac{\mathrm{d}x}{\mathrm{d}t}\,dt=|S_1|-|S_2|.$$
I hope the notation I have used hasn't confused/scared you too much (If it has, don't worry, as I'm pretty sure I would have found it a bit scary at your level). It definitely would have been a lot easier to show you with a picture.

### ...and I've just realised I

...and I've just realised I wrote $\sin t+t\cos t$ when I meant $\sin t-t\cos t$. I should probably just go to bed now.

### Thanks - understood the

Thanks - understood the explanation; although the answer is correct in the mark scheme, I think there is a sign error in the last line of the solution for Q8 - the mark scheme should subtract $|S_2|$ as you said.

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