Thread: area bounded by ∫∫ sin(theta)dA

3. Re: area bounded by ∫∫ sin(theta)dA

it says the area outside the circle radius 2, and inside the cardoid.

your picture shows the inside of the circle shaded.

Further you say in the first octant but your picture shows the entire first quadrant included.

Clean things up please.

4. Re: area bounded by ∫∫ sin(theta)dA Originally Posted by romsek it says the area outside the circle radius 2, and inside the cardoid.

your picture shows the inside of the circle shaded.

Further you say in the first octant but your picture shows the entire first quadrant included.

Clean things up please.
after correction , my ans is 3 , not 8/3 , why am i wrong ? 5. Re: area bounded by ∫∫ sin(theta)dA

do you know what an octant is ?

6. Re: area bounded by ∫∫ sin(theta)dA Originally Posted by romsek do you know what an octant is ?
first octant = 0 to 90 degree , right ?

7. Re: area bounded by ∫∫ sin(theta)dA Originally Posted by xl5899 first octant = 0 to 90 degree , right ?
that's quadrant, and octant is $\dfrac 1 8$ the unit circle.

8. Re: area bounded by ∫∫ sin(theta)dA Originally Posted by romsek do you know what an octant is ?
u mean octant = pi / 8 ???
If so , then the notes at b is wrong , right ?
the shaded area is from 0 to pi / 2 , right ? the correct area should be (81/2) / 8 ? 9. Re: area bounded by ∫∫ sin(theta)dA Originally Posted by xl5899 u mean octant = pi / 8 ???
An octant is $\displaystyle \frac{2\pi}{8}=\frac{\pi}{4}$ radians.

10. Re: area bounded by ∫∫ sin(theta)dA Originally Posted by rubikwizard An octant is $\displaystyle \frac{2\pi}{8}=\frac{\pi}{4}$ radians.
an octant = a quadrant ?

11. Re: area bounded by ∫∫ sin(theta)dA Originally Posted by xl5899 an octant = a quadrant ?
An octant is half of a quadrant. A quadrant is 90 degrees. An octant is 45 degrees.

12. Re: area bounded by ∫∫ sin(theta)dA

Then, the notes that I uploaded in post #8 is wrong?

13. Re: area bounded by ∫∫ sin(theta)dA

Remember that a full circle is $\displaystyle 2\pi$ radians (360 degrees).

In your posts above I think you are using $\displaystyle \pi$ radians for a full circle.

An octant is what you get when you divide a full circle into 8 parts.

14. Re: area bounded by ∫∫ sin(theta)dA Originally Posted by rubikwizard Remember that a full circle is $\displaystyle 2\pi$ radians (360 degrees).

In your posts above I think you are using $\displaystyle \pi$ radians for a full circle.

An octant is what you get when you divide a full circle into 8 parts.
so , the notes is wrong ?

15. Re: area bounded by ∫∫ sin(theta)dA

I haven't gone through the notes you gave in detail, but the replies above from romsek and myself are correct - an octant is half of a quadrant i.e. an octant = 45 degrees, or $\displaystyle \frac{2\pi}{8}$ radians. I still think that from what you have posted you are using a full circle to be $\displaystyle \pi$ radians instead of $\displaystyle 2\pi$ radians.

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