Area between Curves: Find the Bounded Region

  • Thread starter Thread starter flyers
  • Start date Start date
  • Tags Tags
    Area Curves
flyers
Messages
28
Reaction score
0

Homework Statement



Find the area bounded by the curves given by the graphs x = 4 - y^2 and x = y^2 - 2y


The Attempt at a Solution



I don't know how I can integrate these curves as they are not functions. Can someone tell me to get started on this problem?

Thank you
 
Physics news on Phys.org
flyers said:

Homework Statement



Find the area bounded by the curves given by the graphs x = 4 - y^2 and x = y^2 - 2y


The Attempt at a Solution



I don't know how I can integrate these curves as they are not functions. Can someone tell me to get started on this problem?

Thank you

They most certainly are functions. They just don't have x as the independent variable like you're used to. It matters little, you can still get it done.

Integrating is the same no matter the variable of integration...
 
flyers said:

Homework Statement



Find the area bounded by the curves given by the graphs x = 4 - y^2 and x = y^2 - 2y


The Attempt at a Solution



I don't know how I can integrate these curves as they are not functions. Can someone tell me to get started on this problem?

Thank you
They aren't functions, but they are curves, and they define a bounded region. Start by graphing them and finding where the two curves intersect, and then figure out what your typical area element looks like - i.e., horizontal strip of vertical strip- and its dimensions.
 
Mark44 said:
They aren't functions, but they are curves, and they define a bounded region. Start by graphing them and finding where the two curves intersect, and then figure out what your typical area element looks like - i.e., horizontal strip of vertical strip- and its dimensions.

Actually, they are functions. They're just functions of y, not the functions of x most people are used to.
 
you could also just solve for y with those two equations and you would have what you are use to. Just think of those as respect to the y-axis and not the x. So the region you are measuring is taking the integral of the y dy instead of x dx, still the same just a different reference.
 
Char. Limit said:
Actually, they are functions. They're just functions of y, not the functions of x most people are used to.
True, but people generally think of functions where y is the dependent variable and x is the independent variable, out of force of habit. In that sense, the equations don't represent functions.

In any case, the important thing is that the OP should graph both equations to find the region whose area is to be found.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top