Question

what is the figures showing an element of area and what is the area bounded by the curves y^2=4ax and x^2=4ay?

Answer 1

Can you find points of intersection between those two relations?

Answer 2

0 and 0?

Answer 3

(0,0) is one point of intersection. You'll need another.

Hint: There is symmetry about!

Answer 4

Can you solve it for me? :)

Answer 5

Yes.

When was the last time you factored a "Difference of Cubes"? You need to have your algebra up to speed!

Answer 6

aww
last 2 years?

Answer 7

Fair enough. It's time to do it again!

Given these: y^2=4ax and x^2=4ay
I recommend this solution:

x^2=4ay

Assuming everything is positive, there is no harm in squaring things.

\(x^{4}=(4ay)^{2} = 16a^{2}y^{2}\)

Are you bold enough to ponder x^4?! So far so good?

Answer 8

y^2 = 4ax
y = ±2√(ax)
x^2 = 4ay
y = x^2/4a
A = ∫ 2√(ax) - x^2/(4a) dx [0,4a]
= 4/3 * x √(ax) - x^3/(12a) [0,4a]
= 4/3 * 4a * 2a - (64a^3)/(12a)
= (32a^2)/3 - (16a^2)/3
= (16a^2)/3

Answer 9

Where did you get the limit of 4a?

Answer 10

is it right? it is the same answer in the book

Answer 11

Answer my question. Magic won't help you on an exam.

Answer 12

in the 2nd equation?

Answer 13

Nope. It's not there. You have it, but you have not proven it. How can you PROVE that 4a is the correct upper limit?