OpenStudy (dls):

Calculate Area bounded between:

5 years ago
OpenStudy (dls):

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

5 years ago
OpenStudy (anonymous):

So integrate.

5 years ago
OpenStudy (dls):

@ganeshie8

5 years ago
ganeshie8 (ganeshie8):

@hartnn

5 years ago
ganeshie8 (ganeshie8):

can we simply integrate x=0 to 2a ?

5 years ago
ganeshie8 (ganeshie8):

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

5 years ago
ganeshie8 (ganeshie8):

not getting ideas

5 years ago
hartnn (hartnn):

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

5 years ago
ganeshie8 (ganeshie8):

ohk..

5 years ago
hartnn (hartnn):

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

5 years ago
ganeshie8 (ganeshie8):

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

5 years ago
ganeshie8 (ganeshie8):

5 years ago
ganeshie8 (ganeshie8):

a=1

5 years ago
hartnn (hartnn):

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

5 years ago
ganeshie8 (ganeshie8):

where to go frm here

5 years ago
ganeshie8 (ganeshie8):

wolfram says Standard computation time exceeded...

5 years ago
hartnn (hartnn): 5 years ago
hartnn (hartnn):

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

5 years ago
ganeshie8 (ganeshie8):

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

5 years ago
ganeshie8 (ganeshie8):

5 years ago
ganeshie8 (ganeshie8):

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

5 years ago
OpenStudy (dls):

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

5 years ago
OpenStudy (dls):

that was the main thing......

5 years ago
ganeshie8 (ganeshie8):

yeah, but its easy to visualize this particular curve

5 years ago
ganeshie8 (ganeshie8):

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

5 years ago
OpenStudy (dls):

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

5 years ago
OpenStudy (dls):

i can hardly imagine where "pi" came from

5 years ago
ganeshie8 (ganeshie8):

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

5 years ago
OpenStudy (dls):

why are we subbing that ? :|

5 years ago
ganeshie8 (ganeshie8):

wolfram did that

5 years ago
OpenStudy (dls):

take minimum help of these things :/

5 years ago
ganeshie8 (ganeshie8):

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

5 years ago
ganeshie8 (ganeshie8):

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, $$\theta ->\pi/2$$ $$dx = 4a \sin \theta \cos \theta d\theta$$ $$\large \int_0^{\pi/2} \sqrt{\frac{(2a\sin^2\theta)^3}{2a-2a\sin^2\theta}}$$

5 years ago