Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

how do you find the integral of the following:
integral sqrt 9-h^2 limits 0 to 3

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

how do you find the integral of the following:
integral sqrt 9-h^2 limits 0 to 3

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Are you learning about trig substitutions or polar coordinates?

- anonymous

hey i have a test and i am review for it, its trig sunsitution

- anonymous

substitute = 3 sin x

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

Since in the limit of integral,0a=0
h=3=3sin(a)=>a=pi/2
thus integral becomes
limit 0 to 1 9cos^2(a)da
put cos^2(a)=(2cos^(2a)-1)/2
then integrate

- anonymous

Okay, since we have sqrt(a^2 - u^2) form, we have to use a sin t substitution. Let u = asint
\[h = 3\sin \theta\]
You should find that sine of theta is equal to h / 3. Use SOHCAHTOA to draw a right triangle. One leg should be h and the hypotenuse should be 3. Find the other leg using the pythagorean theorem to get sqrt(9 - h^2). Take the cosine of the angle
\[\cos \theta=\sqrt {9-h^2}/3\]
Multiply by 3 on both sides to get cosine(theta)/3 to use as a substitution. Substitute that for sqrt(9 - h^2) in the integral. Now go back to the h = 3sin(theta) equation and take the derivative to get dh = 3cos(theta). Substitute 3cos(theta) for dh to get the integral:
\[\int\limits_{\theta(0)}^{\theta(3)}\cos \theta /3*3\cos \theta d \theta\]
\[\int\limits_{\theta (0)}^{\theta (3)}\cos ^2 \theta d \theta\]
You'll then have to integrate that and back-substitute.

- anonymous

you can also draw a circle with radius 9 and conclude that the area is just one-fourth of the circle...

- anonymous

radius 3, silly me

- anonymous

okay thank you for the explanation but this question but this question is related to another problem u think u can help me

- anonymous

ok

- anonymous

can u please click on the link to see the picture i have to find the area by writing the definite integral and evaluate it http://www.twiddla.com/504415

- anonymous

What area do you need? The area between the two chords?

- anonymous

no thats the strip and u have to use the strip to find the area of the entire circle

- anonymous

you would have to first write riemann sum then the definite integral

Looking for something else?

Not the answer you are looking for? Search for more explanations.