Need Some Math Help
Find the area of the region between the curve x=y^(2/3) and the curve x=2-y^4 . I'm having trouble setting up my integral since I don't exactly see the region I'm suppose to find when I plot the graphs.
November 10, 2012
7 Comments • Newest first
[quote=Watermelon]When you find the points, for example: -2 , 2
You can choose 1 because it is in between -2 and 2.
f1(1) = 2x + 1 = 2 * 1 + 1 = 3
f2(1) = 3x + 1 = 3 * 1 + 1 = 4
f2 > f1
y2 - y1[/quote]
Thanks, that was the main problem that was bothering me.
When you find the points, for example: -2 , 2
You can choose 1 because it is in between -2 and 2.
f1(1) = 2x + 1 = 2 * 1 + 1 = 3
f2(1) = 3x + 1 = 3 * 1 + 1 = 4
f2 > f1
y2 - y1
[quote=LampShadow]Like what watermelon said:
- Set the equations equal to each other, then find the points where they intersect
- Set up the integrals for both equations, subtracting the "lower" equation from the "upper" one. And then use the y points you found before^ as the bounds.
- Solve
That's pretty much it
*edit: if the x and y switch is bothering you, just ignore it. Treat the x variable on the "y axis" and the y variable on the "x axis", then solve it like you would normally. Instead of x boundaries in the integral, use the y points as the bounds[/quote]
How do I know which is the upper and which one is the lower?
Like what watermelon said:
- Set the equations equal to each other, then find the points where they intersect
- Set up the integrals for both equations, subtracting the "lower" equation from the "upper" one. And then use the y points you found before^ as the bounds.
- Solve
That's pretty much it
*edit: if the x and y switch is bothering you, just ignore it. Treat the x variable on the "y axis" and the y variable on the "x axis", then solve it like you would normally. Instead of x boundaries in the integral, use the y points as the bounds
Solve the spots where the curves cut each other and integrate y1 - y2 with those spots.
[quote=robino911]Oh, my.
The @Mathematician: Needs help with math.[/quote]
Maybe we should call the @Mathematician , he might help.
Oh, my.
The @Mathematician: Needs help with math.