Thread: Problem 15: Area between two curves

1. Problem 15: Area between two curves

Let R be the region in the upper half plane between the curves $2x^4 + y^4 + y = 2$ and $x^4 + 8y^4 + y = 1$.

Find: $\iint _R \frac{1}{(x^2 + 1)y}~dy~dx$

-Dan

2. Re: Problem 15: Area between two curves

Hint: Is there a relationship between the two boundary curves?

-Dan

3. Re: Problem 15: Area between two curves

Let's start by looking at the hint and describe the curves.
1) $2x^4 + y^4 + y = 2$

2) $x^4 + 8y^4 + y = 1$

Notice that equation 2) can be rewritten as $2x^4 + (2y)^4 + (2y) = 2$ so equation 2) is just equation 1) with a vertical stretch factor of 2.

I am going to represent equation 1) as y = f(x) and equation 2) as y = 2 f(x).

A couple of important facts about curve 1). Note that f(-x) = f(x), so f(x) is an even function. Also note that f(x) has two real zeros: $x = \pm 1$. And finally, if we look at y = f(x) in quadrant 1 we find that f(x) is monotonically decreasing on [0, 1], so f(x) is one to one in Quadrant 1. (And thus is a function on [-1, 1].) Similar comments obviously apply to y = 2 f(x), and it is trivial to show that the only intersection points of equations 1) and 2) are at $x = \pm 1$.

Our region of integration can thus be characterized as (2 f(x) - f(x))dx integrated over the interval [-1, 1].

Thus
$\iint _{R} \frac{1}{y(x^2 + 1)}~dy~dx = \int _{-1}^{1} \int _{f(x)} ^{2f(x)} \frac{1}{(x^2 + 1)y}~dy~dx$

The integration is now simple. I'll sketch it briefly:
$\iint _{R} \frac{1}{(x^2 + 1)y}~dy~dx = \int _{-1}^{1} \int _{f(x)} ^{2f(x)} \frac{1}{(x^2 + 1)y}~dy~dx$

$= \int_{-1}^{1} \frac{1}{x^2 + 1} \left [ \int_{f(x)}^{2f(x)} \frac{1}{y}~dy \right ] dx$

$= \int_{-1}^{1} \frac{1}{x^2 + 1} \left ( ln(2 f(x)) - ln(f(x) \right ) ~dx$

$= ln(2) \int_{-1}^{1} \frac{1}{x^2 + 1}~dx$

$= \ln(2) \left ( tan^{-1}(1) - tan^{-1}(-1) \right )$

$= \frac{\pi}{2} ln(2)$

-Dan

4. Re: Problem 15: Area between two curves

There are four points of intersection:

$(-1,0), (1,0), \left( -\sqrt[4]{1+7\cdot 15^{-4/3} }, -15^{-1/3} \right) \approx (-1.04428,-0.405480), \left( \sqrt[4]{1+7\cdot 15^{-4/3} }, -15^{-1/3} \right)\approx (1.04428,-0.405480)$

Does your solution account for all four points of intersection?

Edit: Never mind. I missed the part that said that $R$ is restricted to the upper half-plane.

5. Re: Problem 15: Area between two curves

Originally Posted by SlipEternal
There are four points of intersection:

$(-1,0), (1,0), \left( -\sqrt[4]{1+7\cdot 15^{-4/3} }, -15^{-1/3} \right) \approx (-1.04428,-0.405480), \left( \sqrt[4]{1+7\cdot 15^{-4/3} }, -15^{-1/3} \right)\approx (1.04428,-0.405480)$

Does your solution account for all four points of intersection?

Edit: Never mind. I missed the part that said that $R$ is restricted to the upper half-plane.
Actually it's good that you pointed that out. I wasn't looking at the lower half plane very carefully. I had simply assumed there were no other crossings and apparently there are. Thanks for the comment.

-Dan