Find the area bounded by an ellipse x2/16+y2/9=1 using definite integral

Question

X²/16+y²/9=1

Answer 1

X²/16+y²/9=1

The ellipse is centered in origin and its area consists of 4 equal parts, each in one of four quadrants. Therefore, we find area of one pars and multiply by 4.

So y=3√(1-x²/16). 

When y=9, x=-4  or 4. Since we taking one pars, the integration limits would be 0 and 4

S=4∫₀⁴3√(1-x²/16)dx

=|substitute  x=4sin(t), dx=4cos(t)dt. Since 0≤x≤4,  0≤t≤π/2|=

=4*3*4∫₀^π/2 cos²(t)dt=2*12∫₀^π/2(1+cos(2t)dx=2*12(t+(1/2)sin(2t))|₀^π/2=12π.