Question

Halp

Answer 1

\[\int\limits_{0}^{a} x^2\sqrt{a^2-x^2}dx\]

Answer 2

correct just funny way to say help

Answer 3

@Nnesha

Answer 4

@ganeshie8

Answer 5

This would also be a problem of solving the integral using trig substituion

Answer 6

try \(\large x = a\sin u\)

Answer 7

I have and that would make dx = a cos u du and if i change the limits of integration when x=a than u = pi/2 and if i change x =0 then u = pi and I got

Then i solved for what the squareroot portion would equal if i were to insert the x into it and go a cos u out of that so I now have

\[\int\limits_{\pi/2}^{\pi}a^3\sin ^2u \cos u du\]

This is where I am stuck

Answer 8

integrand should be \(\large a^4 \sin^2u \cos^2u\)
check again

Answer 9

well i put x = a sin u in where we got x^2 and got

\[a^2\sin^2u\]

then i solved the square root portion

\[\sqrt{a^2-x^2} when x = a \sin u\]
\[\sqrt{a^2-a^2\sin^2u}\]
\[\sqrt{a^2}\sqrt{1-\sin^2u}\]
\[a \sqrt{\cos^2u}\]
\[a \cos u\]

bring it together

\[(a^2\sin^2u)(a \cos u)( a \cos u)\]

OK now I see where I got that wrong