# Find the area bounded by the parametric curve…

Find the area bounded by the parametric curve $x = \cos(t)$, $y = e^t, 0 < t < \pi/2$, and the lines $y = 1$ and $x = 0$.

I do not even know where to start with this problem. I know that I need to draw a graph, but that's all I know. Thanks for the help!

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How would I set up the definite integral? –  Ryan Nov 2 '12 at 4:41
when you say and the lines $y=1$ and $x=0$, do you mean "at" the lines $y=1$ and $x=0$ –  boby Nov 2 '12 at 4:44

## 2 Answers

If $x=\cos(t)$, then $t=\arccos(x)$, at least when $0\leq t \leq \pi/2$. Thus, your curve can be expressed as $y=e^{\arccos(x)}$, which you can integrate over $[0,1]$.

I certainly agree that a picture like the one below is useful. You could plot both the parametric plot and the function plot using a tool like WolframAlpha. It then becomes apparent that you should subtract 1 from the integral. Of course, you can also ask WolframAlpha for help with the integral, which might be a bit tricky.

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why would you need to draw anything? evaluate the definite integral –  boby Nov 2 '12 at 4:36
@boby: What would the be the a and b values for the integral? –  Ryan Nov 2 '12 at 4:38

To finish the resolution of this problem, we would continue from the graph shown by Mark McClure : the bounding curve is traced out from $\ ( \ \cos \ 0 \ , \ e^0 \ ) \ = \ ( \ 1 \ , \ 1 \ ) \$ to $\ ( \ \cos \frac{\pi}{2} \ , \ e^{\pi / 2} \ ) \ = \ ( \ 0 \ , \ e^{\pi / 2} \ ) \$ . The region for which we wish to find the area is thus perched atop a unit square:

We would set up the area integral as $\ \int_0^1 \ y(x) \ - \ 1 \ \ dx \$ , but to carry this out parametrically , we need to account for the curve being traced "backwards" (from right to left):

$$\int_0^1 \ y(x) \ - \ 1 \ \ dx \ \ \rightarrow \ \ \left[ \ \int^0_{\pi / 2} \ y(t) \ \frac{dx}{dt} \ \ dt \ \right] \ - \ 1 \ \ = \ \ \left[ \ \int^0_{\pi / 2} \ e^t \ ( - \sin \ t \ \ dt ) \ \right] \ - \ 1$$

$$= \ \ \left[ \ \int_0^{\pi / 2} e^t \ \sin \ t \ \ dt \ \right] \ - \ 1 \ = \ \ \frac{1}{2} e^t \ ( \ \sin \ t \ - \ \cos \ t \ ) \vert_0^{\pi / 2} \ - \ 1$$

[having integrated by parts twice]

$$= \ \ \frac{1}{2} \left[ \ e^{\pi / 2} \ ( \ \sin \frac{\pi}{2} \ - \ \cos \frac{\pi}{2} \ ) \ - \ e^0 \ ( \ \sin \ 0 \ - \ \cos \ 0 \ ) \ \right] \ - \ 1$$

$$= \ \frac{1}{2} \left[ \ e^{\pi / 2} \ ( \ 1 \ - \ 0 \ ) \ - \ 1 \ ( \ 0 \ - \ 1\ ) \ \right] \ - \ 1 \ = \ \frac{1}{2} \left( \ e^{\pi / 2} \ + \ 1 \ \right) \ - \ 1$$

$$= \ \frac{1}{2} \left( \ e^{\pi / 2} \ - \ 1 \ \right) \ \approx \ 1.905 \ \ .$$

Despite the curve not (quite) being a straight line, the area is estimated very well by a triangle of appropriate dimensions: $\ \frac{1}{2} \ \cdot \ ( \ 1 - 0 \ ) \ \cdot \ ( \ e^{\pi / 2} \ - \ 1 \ ) \$ .

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