Calculate Area bounded between:

Question

dls · Answer

Curve $\LARGE y^2(2a-x)=x^3$ and line $\LARGE x=2a$

anonymous · Answer

So integrate.

dls · Answer

@ganeshie8

ganeshie8 · Answer

@hartnn

ganeshie8 · Answer

can we simply integrate x=0 to 2a ?

ganeshie8 · Answer

$\large \int_0^{2a} \sqrt{\frac{x^3}{2a-x}} dx$

ganeshie8 · Answer

not getting ideas

hartnn · Answer

no, i don't think so
i was thinking of trigonometric substitution first..to get polar forms of equation of curve

ganeshie8 · Answer

ohk..

hartnn · Answer

i am not able to graph/plot the curve...

ganeshie8 · Answer

yeah, a is there... we will get family of curves

ganeshie8 · Answer

attachment

ganeshie8 · Answer

a=1

hartnn · Answer

then $\Large \large \int_0^{2a} \sqrt{\dfrac{x^3}{2a-x}} dx$
is correct

ganeshie8 · Answer

where to go frm here

ganeshie8 · Answer

wolfram says
Standard computation time exceeded...

hartnn · Answer

http://www.wolframalpha.com/input/?i=integral+sqrt+%28+x%5E3%2F+%282a-x%29%29

hartnn · Answer

i can't imagine from where did he get this monstrous question...

ganeshie8 · Answer

oh wolfram solution looks neat, its subbing $x = 2a sin^2\theta$

ganeshie8 · Answer

attachment

ganeshie8 · Answer

thanks @hartnn i entered definite integral initially and it was giving up... http://www.wolframalpha.com/input/?i=int+0+to+2a+sqrt%28x%5E3%2F%282a-x%29%29

dls · Answer

i won't get internet connection to plot the curve .-.

dls · Answer

that was the main thing......

ganeshie8 · Answer

yeah, but its easy to visualize this particular curve

ganeshie8 · Answer

$\large y = \sqrt{\frac{x^3}{2a-x}}  $

dls · Answer

Aapke options,apki screen par
$$\LARGE 3 \pi a ^2$$
$$\LARGE \frac{3 \pi a ^2 }{2}$$
$$\LARGE \frac{3 \pi a^2}{4}$$
$$\LARGE None ~$$

dls · Answer

i can hardly imagine where "pi" came from

ganeshie8 · Answer

pi comes in limits, after we sub x = 2a sin^2

dls · Answer

why are we subbing that ? :|

ganeshie8 · Answer

wolfram did that

dls · Answer

take minimum help of these things :/

ganeshie8 · Answer

wat things, wolfram solution doesnt make sense to u yet ?

ganeshie8 · Answer

see if below is any clear :-

$\large \int_0^{2a} \sqrt{\frac{x^3}{2a-x}}$

say $x = 2a \sin^2 \theta$

x->0, $\theta ->0$
x->2a, $\theta ->\pi/2$

$dx = 4a \sin \theta \cos \theta d\theta$


$\large \int_0^{\pi/2} \sqrt{\frac{(2a\sin^2\theta)^3}{2a-2a\sin^2\theta}}$