Question

the curve y^2=x(x-n)^2, when n is a psitive integer,encloses a loop. find exact area enclosed by this loop for n=1,2,3 and 4

Answer 1

try:
\[\int\limits_{0}^{n} -2\sqrt{x}(x-n) dx\]

Answer 2

Above times two.

Answer 3

why's that?

Answer 4

...shoot it was y^2=x^2(x-n)

Answer 5

@Algebraic! It only gives area with x axis.
The area was both upwards and downwards.

Answer 6

sqyaure root of x^2 is x

Answer 7

http://www.wolframalpha.com/input/?i=y%3Dsqrtx+%28x-1%29

Answer 8

y= +/- sqrt(x)(x-n)

Answer 9

nice sid

Answer 10

@Algebraic! Even if its +- You are integrating it just once right?

Answer 11

I thought the y^2 cancxeled out the x^2?

Answer 12

you integrate the top portion, since it's positive, then you double it.

Answer 13

so...Its x square root of x-n)?

Answer 14

or, alternately, integrate the bottom portion times -2

Answer 15

Precisely.

Answer 16

l2read

Answer 17

ok...

the formula for area as a function of n...?\[A=2\int\limits_{0}^{?}\]

Answer 18

\[A=\int\limits_{0}^{n}-2x \sqrt{(x-n)}\]

Answer 19

??

Answer 20

note: you said
..shoot it was y^2=x^2(x-n)

But this does not form a loop.
see http://www.wolframalpha.com/input/?i=y%5E2%3Dx%5E2%28x-n%29%2C+where+n%3D2

On the other hand, the original equation does form a loop
http://www.wolframalpha.com/input/?i=y%5E2%3Dx%28x-n%29%5E2%2C+where+n%3D2

Answer 21

you are right