2
$\begingroup$

I have to find an area using definite integral, where the figure is restricted by two functions, stated in title. Can someone explain how to do that?

Here is my try:

So I equal the expressions to find intersections, - $$\frac{a^3}{a^2+x^2}=\frac{x^2}{2a}$$

$$\frac{2a^4}{a^2+x^2}=x^2$$

$$2a^4=a^2x^2+x^4$$ therefore $${2a^4-a^2x^2-x^4}=0$$

$$x=\frac{a^2 \pm \sqrt{a^4-4 \cdot (-1) \cdot 2a^4}}{{-2}}=\frac{a^2 \pm \sqrt{9a^4}}{-2}$$ $$x_1=\frac{4a^2}{-2}=-2a^2$$ and $$x_2=a^2$$

I don't know what to do next, and doubt that my calculations are correct.

$\endgroup$
  • $\begingroup$ Did you mean $y=\dfrac{a^3}{a^2+x^2}$? What have you tried? Did you at least find out where the two curves intersect? You need to include in your question what you already know about the problem, what you have tried, etc, or your question will continue to be voted down. $\endgroup$ – John Wayland Bales Jun 22 at 0:28
  • $\begingroup$ @JohnWaylandBales I have edited my post, where I try to find an intersection between two curves. $\endgroup$ – Ieva Brakmane Jun 22 at 15:18
  • 1
    $\begingroup$ @IevaBrakmane you can find the intersections with the second degree equation formula because you are getting a biquadratic equation. $\endgroup$ – dcolazin Jun 22 at 15:35
  • 1
    $\begingroup$ You have written $a^2x^2+x^4=2a^4-a^2x^2-x^4$ which is not true. That equal sign should be a therefore or a carriage return. What you have are two equations which are equivalent. But you cannot use an $"="$ sign to indicate the equality of two equations, only the equality of two expressions. You should say $2a^4=a^2x^2+x^4$ therefore $2a^4-a^2x^2-x^4=0$. You can solve that second equation for $x^2$ using the quadratic formula. $\endgroup$ – John Wayland Bales Jun 23 at 6:19
  • $\begingroup$ @JohnWaylandBales I really doubt that my computations will be correct, but here it goes: $x=\frac{a^2 \pm \sqrt{a^4-4 \cdot (-1) \cdot 2a^4}}{{-2}}=\frac{a^2 \pm \sqrt{9a^4}}{-2}$ $x_1=\frac{4a^2}{-2}=-2a^2$ and $x_2=a^2$ $\endgroup$ – Ieva Brakmane Jun 24 at 12:22
1
$\begingroup$

When you are applying the quadratic equation, you are solving for $x^2$, not $x$ since this is a fourth degree polynomial equation which is quadratic in form.

The first solution is not possible because $x_1^2 $cannot be negative.

So that leaves the second solution $x_2^2=a^2$.

So you get two solutions for $x$.

You get $x=\pm a$.

So to get the area of your figure you must subtract the lesser of the two functions from the greater and integrate the difference over the interval $[−a,a]$.

So you must solve the integral

$$ \int_{-a}^a\frac{a^3}{a^2+x^2}-\frac{1}{2a} x^2\,dx$$

The first term will involve the arctangent function and the second term is a polynomial.

NOTE: Since the region is symmetric with respect to the $y$-axis, you can find the area by doubling the integral from $0$ to $a$, simplifying the process somewhat.

$\endgroup$
  • $\begingroup$ So to determine which of those two functions are greater you simply plug in $a$ whichever real number except $0$? So I looked up at integral calculator, and to integrate the first part $\frac{a^3}{a^2+x^2}$ they use substitution $u=\frac{x}{a}$. It works here because in the end we get the derivative of arctan. But in similar problems, how do I know when to make such a substitution? Then when I integrated the result for me it was $\frac{a^2\pi}{2}-\frac{a^3}{3}$ Is it correct? $\endgroup$ – Ieva Brakmane Jun 26 at 11:34
  • $\begingroup$ Almost correct, except that should be $\dfrac{a^2}{3}$ instead of $\dfrac{a^3}{3}$. If you are taking a calculus class, you should have studied integrals involving the inverse trigonometric functions. Have you not studied them? $\endgroup$ – John Wayland Bales Jun 26 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.