Question

I need to find the integral from -6 to 0 for sqrt (36-x^2). Have been trying for 20 minutes and can't get right answer. Need to know how to start because obviously I am not starting it right.

Answer 1

Take a substitution of x=6sin(t) I think u can carry rest..if not tell me

Answer 2

Where are you getting the sin from?

Answer 3

\[\int\limits_{-6}^{0}\sqrt{(36-x^{2}})dx\]

Answer 4

\[\int\limits_{-6}^{0}\sqrt{36-x^2}dx\]Let x=6sint
then, dx=6cost dt

Answer 5

\[\int\limits_{-\pi/2}^{0}\sqrt{36-36\sin^2t}6\cos tdt\]

Answer 6

now you have 36(1-sin^2 (t) in the root sign. sin^2(t)+cos^2(t)=1 to eliminate the root. go from there...

Answer 7

http://www.wolframalpha.com/input/?i=integrate+sqrt+%2836-x%5E2%29.
Click on "Show steps"

Answer 8

How do I mark that my question was given a good answer?

Answer 9

click good answer on one of us to give a medal

Answer 10

\[\int\limits_{-\Pi/2}^{0}\sqrt{6^2-6^2\sin^2(t)}\cos(t)dt\]
\[\int\limits_{-\Pi/2}^{0}6\sqrt{1-\sin^2(t)}\cos(t)dt\]
\[\int\limits_{-\Pi/2}^{0}6\sqrt{\cos^2t}\cos(t)dt\]
\[\int\limits_{-\Pi/2}^{0}6{\cos(t)}\cos(t)dt\]
\[\int\limits\limits_{-\Pi/2}^{0}{6\cos^2(t)dt}\]
\[\int\limits\limits_{-\Pi/2}^{0}6{(\cos(2t)+1)/2}dt\]
\[[\sin(2t)/2 +t](0,-\Pi/2)\]
\[\Pi/2\]

Answer 11

Where do I do that ? It says that I need to log in but it says that I am already logged in.

Answer 12

weird. try leaving the site and coming back.

Answer 13

Got it! Thanks for everyones help. I see what I did wrong now.

Answer 14

Medals given!!