Question

Use integration and trigonometric substitution to find the area enclosed by the ellipse: (x^2)/16 + (y^2)/25 = 1

Answer 1

Or ...

we could just use the fact that the area of
x^2 / a^2 + y^2 / b^2 = 1
is π(ab)
for ours: a = 4 and b = 5, so the area = 20π units^2

but , if you insist on integration ....
Because of the nice symmetry
Area = 4 ∫ y dx from 0 to 4

we get y^2 / 25 = 1 - x^2/16
y^2 = 25(16 - x^2)/16
y = (5/4)√( 16 - x^2)

Let x = 4sinθ
dx = 4cosθ dθ

y = (5/4)√(16 - 16sin^ θ) = (5/4)(4)√(1 - cos^2 θ)
= 5√(sin^2 θ) = 5sinθ

area = 4 ∫5 sinθ (4cosθ) dθ ,

when x=0, 0 = 4sinθ, θ=0
when x = 4, 4 = 4sinθ, θ = π/2

= 40 ∫ 2sinθcosθ dθ form 0 to π/2
= 40[ sin^2 θ] from 0 to π/2
= 40( (√2/2)^2 - 0)
= 20, I appear to have lost the value of π, should be 20π
but I checked and checked and can't find my error.

Answer 2

found my error!

I seem to have lost concentration in :
y = (5/4)√(16 - 16sin^ θ) = (5/4)(4)√(1 - cos^2 θ)
= 5√(sin^2 θ) = 5sinθ

of course that should have been
y = (5/4)√(16 - 16sin^ θ) = (5/4)(4)√(1 - sin^2 θ)
= 5√(cos^2 θ) = 5cosθ

so ...
area = 4 ∫5 cosθ (4cosθ) dθ
= 80 ∫ cos^2 θ dθ from 0 to π/2

the rest of my calculations are now garbage, since we have to
integrate cos^2 θ

arhhhgg, don't feel like doing that, since only about 8 people have even
looked at this question, and it feels like I am doing this just for my own
fun.

Trust me, on my scribble paper I did get 20π

Answer 3

or, let
x = 4cosθ, dx = -4sinθ dθ
y = 5sinθ, dy = 5cosθ dθ

4 ∫[0,4] y dx = 4 ∫[0,π/2] 5sinθ (-4cosθ dθ)
= -40∫[0,π/2] sin2θ dθ
= -40 (-1/2 cos2θ) [0,π/2]
= 20π