a^2y^2=x^2(x^2-a^2) find out the area using definite integral

Question

plz solution of the question

Comments

For a definite integral you must specify limits.

Answer 1

Let's assume the area required is between x=p and x=q. This curve has open arms, which means that some areas will be infinite. This imposes constraints on the choice of p and q. p²≥a² fulfils one constraint. Also q>p. Consider the integral _a∫^qydx, where y=x√(x²/a²-1)). This is the result of taking the square root of both sides and making y a function of x.

Let x=asecθ, so dx=asecθtanθdθ. When x=a, secθ=1⇒θ=0. When x=q, secθ=q/a, cosθ=a/q, θ=arccos(a/q), tanθ=√(q²/a²-1)=√(q²-a²)/a.

Let J=∫x√(x²/a²-1))dx=a∫secθtan²θdθ=a∫sec³θdθ-a∫secθdθ.

Let I=∫sec³θdθ=∫secθ(sec²dθ). Let u=secθ, du=secθtanθ, dv=sec²θdθ, v=tanθ.

I=uv-∫vdu=secθtanθ-∫secθtan²θdθ=secθtanθ-∫sec³θdθ+∫secθdθ,

I=secθtanθ-I+∫secθdθ;

secθ=secθ(secθ+tanθ)/(secθ+tanθ)=(sec²θ+secθtanθ)/(tanθ+secθ),

2I=secθtanθ+ln|tanθ+secθ|, I=½secθtanθ+½ln|tanθ+secθ|.

J=a(½secθtanθ+½ln|tanθ+secθ|)-aln|tanθ+secθ|=½a(secθtanθ-ln|tanθ+secθ|).

When θ=0, J₀=0; θ=arccos(a/q), J_arccos(a/q)=½(q/a)√(q²-a²)-½aln((√(q²-a²)+q)/a).

The area J=J_arccos(a/q)-J₀=½(q/a)√(q²-a²)-½aln((√(q²-a²)+q)/a) if integration is over the interval [a,q].

However, this is the area above the x-axis. There is an equal area below the x-axis so the total area over the same interval is (q/a)√(q²-a²)-aln((√(q²-a²)+q)/a).