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!

 

[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

 

[lurflurf]

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

 

[lurflurf]

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.