Question

Integral of sqrt(64-x^2)?

Answer 1

\[\int\limits \sqrt{64-x ^{2}}dx\]

Answer 2

Try the trig substitution \(x=8\sin\theta\).

Do you think you can take things from here? :-)

Answer 3

, no why would that help?

Answer 4

oh doh, what @ChristopherToni said

Answer 5

I'm not very good with trig identities :/

Answer 6

try it and see,
it is the trick to get rid of the square root
probably saw this back in trig class and figured you would never use it

Answer 7

ohhhhhhh

Answer 8

Let's see why this would be case:

If \(x=8\sin \theta\), then it follows that \(dx = 8 \cos\theta\,d\theta\).

Furthermore,
\[\begin{aligned}\sqrt{64-x^2} &= \sqrt{64-(8\sin \theta)^2}\\ &= \sqrt{64-64\sin^2\theta}\\ &= \sqrt{64(1-\sin^2\theta)} \\ &= \sqrt{64\cos^2\theta}\\ &= 8\cos\theta\end{aligned}\]

Therefore,
\(\begin{aligned}\int \sqrt{64-x^2}\,dx &= \int 8\cos\theta\cdot8\cos\theta\,d\theta\\ & = 64\int\cos^2\theta\,d\theta \\ &= 32\int 1+\cos(2\theta)\,d\theta \\ &= 32\theta + 16\sin(2\theta)+C\\ &=32\theta+32\sin\theta\cos\theta +C\end{aligned}\)

At this point, you need to convert everything back to \(x\). You now need to use the following triangle to help you:



Do you think you can wrap things up from here? :-)

Answer 9

\[\int\limits \sqrt{64(1-\sin ^{2}\theta)}dx\]

Answer 10

wow

Answer 11

gimme a min

Answer 12

rings a bell now i bet right?

Answer 13

yup :)

Answer 14

actually the original problem was \[\int\limits_{-8}^{8}\sqrt{64-x ^{2}}\]

Answer 15

so i think i'll just change the limits accordingly

Answer 16

Oh wow.

Well, if that's the case, then this is super trivial if you recall what kind of shape \(\sqrt{64-x^2}\) is; then you can use the corresponding geometry formula to find the value of the integral. XD

Answer 17

ohh

Answer 18

wow i am dumb

Answer 19

semicircle with radius 8

Answer 20

:/

Answer 21

oh my all that for the areas of half a circle
you can still do it the other way, but i am going to guess that this comes in the book long before the stupefyingly dull techniques of integration

Answer 22

oh wait it's the average value so the answer is 2pi

Answer 23

at least it was good practice

Answer 24

that's all i can say

Answer 25

Yes, good practice for what (not) to look forward to in calculus. XD

Answer 26

:) Thanks a ton, both of you guys! I wish I could give 2 best responses.