## anonymous 5 years ago how do you solve y = 3/ sqrt(16-x^2) ?

1. anonymous

$y=\frac{3}{\sqrt{16+x^2}}$ what are you solving for?

2. anonymous

The area of the region between 0 and 2.

3. anonymous

you are looking for $\int^2_0 \frac{3}{\sqrt{16-x^2}}dx$?

4. anonymous

yes.

5. anonymous

I know that it has to do with the sine identity, but I keep getting the wrong answer using that. My answer is never an angle that is correct

6. amistre64

x = 4sin(t)

7. amistre64

dx = 4cos(t) dt

8. anonymous

antiderivative of $\frac{1}{\sqrt{a^2-x^2}}$ is $sin^{-1}(\frac{x}{a})$

9. amistre64

3*4 cos(t) [S] --------- dt right? 4 cos(t)

10. anonymous

$1/a *\arcsin(x/a) ?$

11. amistre64

[S] 3 dt -> 3t ... what does t revert back to tho? x = 4sin(t) x/4 = sin(t) sin^-1(x/4) = t if i did it right :)

12. amistre64

3 * sin^-1(x/4)

13. anonymous

I pulled out the three and had (1/4)arcsin(x/4) evaluated from 0 to 2

14. anonymous

evaluate at upper limit, get $sin^{-1}(\frac{2}{4})=sin^{-1}(\frac{1}{2})=\frac{\pi}{6}$ plug in 0 get 0 so answer is $\frac{\pi}{6}$

15. anonymous

Did you multiply by the 3?

16. anonymous

oops forgot about the ''3" answer is $\frac{\pi}{2}$ sorry

17. anonymous

So when I get a decimal, how can I revert it make to degrees/radians so that the computer will except it?

18. amistre64

$frac{test}{ing}$

19. amistre64

lol... that didnt work out like i thought

20. anonymous

what's that?

21. amistre64

pi 180 -- * ----- = 90 degrees... 2 pi

22. anonymous

pi/2 is right

23. anonymous

the area is the area. As a function this is a function of numbers, not degrees or radians. It corresponds to the radian measure of the angle, not the degree measure. Always work in 'radians' if you are thinking of angles. But they are just numbers.

24. anonymous

okay

25. anonymous

26. amistre64

youre welcome :)

27. anonymous

I hope this is clear. the area is $\frac{\pi}{2}$ square units.

28. anonymous