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Area of loop

suvadip

Member
Feb 21, 2013
69
I have to find the area of the loop of the curve \(\displaystyle a^4 y^2=x^4(a^2-x^2).\)

I have confusion regarding the shape of the graph the limits of integration.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would first exploit the symmetries of the curve and consider only the first quadrant. Then find the $x$-intercepts to obtain the limits of integration, after which a trigonometric substitution works nicely.

Can you proceed?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
As we can see, there are only even powers of $x$ and $y$, and so we know there is symmetry across both coordinate axes. And so the total area $A$ enclosed will be 4 times the area in the first quadrant. The non-negative $x$-intercepts are found by equating $y$ to zero:

\(\displaystyle 0=x^4\left(a^2-x^2 \right)=x^4(a+x)(a-x)\)

And so we find these intercepts are at:

\(\displaystyle x=0,\,a\)

And so we may state:

\(\displaystyle A(a)=\frac{4}{a^2}\int_0^a x^2\sqrt{a^2-x^2}\,dx\)

At this point, we may consider the substitution:

\(\displaystyle x=a\sin(\theta)\,\therefore\,dx=a\cos(\theta)\)

So, I will now stop at this point to give you a chance to take it from here. :D