Question

Need help to figure out an antiderivative. squareroot(4-x^2)

Answer 1

\[\int\limits_{0}^{b}16\sqrt{4-x ^{2}}\]

Answer 2

if forgot to add the 16

Answer 3

we're working with volumes and we want to find one of the endpoint b that would give a volume of 19.654

Answer 4

i figure that by using the fundamental theorem i could solve for that endpoint

Answer 5

using trig-sub it will work
first, let x=2sin(theta)
what happend next ?

Answer 6

we have
\[16\sqrt{4-4\sin ^{2} \theta}\]

Answer 7

simplify, can be to sqrt(4-4sin^2 theta) = sqrt(4(1-sin^2 theta)) = sqrt(4cos^2 theta)
= 2cos(theta)
note : sin^2 (theta) + cos^2 (theta) = 1

Answer 8

and because x=2sin(theta) (differentiate both side)
so, dx=2cos(theta) d(theta)

Answer 9

\[\int\limits_{0}^{b}\frac{ 16 }{ \sqrt{4-x^2} }\]

Answer 10

i think your question is like this

Answer 11

but i dont see divide at first

Answer 12

03453660 hmm, no its like i posted originally. I'm sure
RadEn: i see what you are saying there, my trig is weak so i couldnt i denitrify that

Answer 13

i meant identify

Answer 14

ok, that problem can be :
int (16*2sin(theta)*2cos(theta) d(theta)
= 32 int (sin(2*theta)) d(theta)

Answer 15

you lost me there, i was just going to find the antiderivative of 2cos(theta) since 16 is a constant

Answer 16

= -16cos(2*theta)
and then from x=2sin(theta), (i want to show in pic)