• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Calc problem (area of parametric curves)

  • Thread starter itzela
  • Start date
34
0
calc 2 problem (area bound by parametric eq.)

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

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

But apparently I'm missing steps... which i'm not aware of! Please help!
 
Last edited:
111
0
try to write t in terms of x and then replace t with what you have found in the other equation...
 
34
0
What i did was i used the formula [tex] Area= \int f(t)g'(t)dt [/tex] and plugged in [tex] f(t) = e^t\ g'(t) = -sint [/tex],
which is how i came to the this: [tex] \int e^t(-sin{t})dt [/tex]
with a bound of [tex] 0\to\pi/2[/tex]

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

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

lurflurf

Homework Helper
2,417
122
itzela said:
What i did was i used the formula [tex] Area= \int f(t)g'(t)dt [/tex] and plugged in [tex] f(t) = e^t\ g'(t) = -sint [/tex],
which is how i came to the this: [tex] \int e^t(-sin{t})dt [/tex]
with a bound of [tex] 0\to\pi/2[/tex]

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

I don't get how the integral in 1. was reached :confused: 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
[tex] =\int_0^1 (y-1)dx [/tex]
or
[tex] =\int_0^1 (e^{{Arccos}(x)}-1)dx [/tex]
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
[tex] = \int_0^\frac{pi}{2} (e^t-1)(-sin(t))dt [/tex]
 
34
0
I'm so sorry... but I still don't understand why the area wouldn't be given by:
[tex] \int_0^\frac{pi}{2} e^t(-sin{t})dt [/tex]

How did arrive to [tex] = \int_0^1 (y-1)dx [/tex]
 

lurflurf

Homework Helper
2,417
122
itzela said:
I'm so sorry... but I still don't understand why the area wouldn't be given by:
[tex] \int_0^\frac{pi}{2} e^t(-sin{t})dt [/tex]

How did arrive to [tex] = \int_0^1 (y-1)dx [/tex]
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

[tex] =\int_a^b (f(x)-g(x))dx [/tex]
This is the same, but in this example f(x)=y g(x)=1 a=0 b=1 so
[tex] =\int_0^1 (y-1)dx [/tex]
 
34
0
Thanks a bunch! That explanation really helped =) I can solve it from here on.
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top