Question

area under curve y=0
x=y^4
y=(2-x)^(1/2)

Answer 1

y=√(2-x)
y² = 2-x
2-y² = x = y⁴

Let u = y²
We find that y = ±1 So \(\displaystyle\int_{-1}^1 (2-y^2) - (y^4) \space dy\)
Can you handle it now?

Answer 2

why cant i integrate it using x
with the bounds of 0 and 2

Answer 3

You can do that, but doing it in terms of y is much easier.

If you want to do it in term of x you have to break up area into two area and find integration separately.

Answer 4

but it is bounded by y=0

Answer 5

D'oh, I mean \[\int_0^1 (2- y^2) - (y^4)\space dy\]

Either way, you will get the same answer.

Answer 6

i got (16/5)(2^(1/4))-(4/3)(2^(1/2))

Answer 7

None. You sure you set it correctly? \[\Large \int_0^1 \sqrt[4]{x} \space dx + \int_1^2 \sqrt{2-x}\space dx\]

Answer 8

i didnt do that
why do i have to split the integral?