# Calc problem (area of parametric curves)

itzela
calc 2 problem (area bound by parametric eq.)

I'm having a problem with this question:
Find the area bounded by the curve $$x=cos{t}\ y= e^t$$,
$$0\geq t\leq\pi/2\$$,
and the lines $$y=1\ x=0$$

... I came up with $$\int e^t(-sin{t})dt$$ from $$0\to\pi/2$$

But apparently I'm missing steps... which i'm not aware of! Please help!

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## Answers and Replies

wisredz
try to write t in terms of x and then replace t with what you have found in the other equation...

itzela
What i did was i used the formula $$Area= \int f(t)g'(t)dt$$ and plugged in $$f(t) = e^t\ g'(t) = -sint$$,
which is how i came to the this: $$\int e^t(-sin{t})dt$$
with a bound of $$0\to\pi/2$$

Although I'm suppose to start of by
1. letting the Area $$= \int (y-1)dx$$
with a bound of $$0\to1$$,
2. And then that would equal to $$\int (e^t - 1)(-sint)dt\$$
with a bound of $$\pi/2\to 0$$
3. Which would result in $$\int e^t sint - sint dt$$
with a bound $$0\to\pi/2$$

I don't get how the integral in 1. was reached or why bounds were changed in each step.... any help would be appreciated

Homework Helper
itzela said:
What i did was i used the formula $$Area= \int f(t)g'(t)dt$$ and plugged in $$f(t) = e^t\ g'(t) = -sint$$,
which is how i came to the this: $$\int e^t(-sin{t})dt$$
with a bound of $$0\to\pi/2$$

Although I'm suppose to start of by
1. letting the Area $$= \int (y-1)dx$$
with a bound of $$0\to1$$,
2. And then that would equal to $$\int (e^t - 1)(-sint)dt\$$
with a bound of $$\pi/2\to 0$$
3. Which would result in $$\int e^t sint - sint dt$$
with a bound $$0\to\pi/2$$

I don't get how the integral in 1. was reached or why bounds were changed in each step.... any help would be appreciated
I think the parametric equations are throwing you of think back to when you first did area problems. Think of y(x)=exp(Arccos(x))
to find the area bounded by x=0 y=1 and y=exp(Arccos(x)) do as you would do then
$$=\int_0^1 (y-1)dx$$
or
$$=\int_0^1 (e^{{Arccos}(x)}-1)dx$$
now do a change of variable
x=cos(t)
dx=-sin(t)dt
0<x<1
becomes
0<t<pi/2
this is how the bounds change
so we have
$$= \int_0^\frac{pi}{2} (e^t-1)(-sin(t))dt$$

itzela
I'm so sorry... but I still don't understand why the area wouldn't be given by:
$$\int_0^\frac{pi}{2} e^t(-sin{t})dt$$

How did arrive to $$= \int_0^1 (y-1)dx$$

Homework Helper
itzela said:
I'm so sorry... but I still don't understand why the area wouldn't be given by:
$$\int_0^\frac{pi}{2} e^t(-sin{t})dt$$

How did arrive to $$= \int_0^1 (y-1)dx$$
It is because you want the area between y and 1. Recall that to find the area between f(x) and g(x) where f(x)>g(x) and x=a, x=b you find

$$=\int_a^b (f(x)-g(x))dx$$
This is the same, but in this example f(x)=y g(x)=1 a=0 b=1 so
$$=\int_0^1 (y-1)dx$$

itzela
Thanks a bunch! That explanation really helped =) I can solve it from here on.