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

1. ## area bounded by ∫∫ sin(theta)dA

Evaluate area bounded by ∫∫ sin(theta)dA , where R is the region outside circle r =2 and inside cardroid r = 2 + 2 cos(theta) , in first octant ....
I have sketched the area , is it correct ?

The ans provided is 4/3 , however i gt 2 . Which part is wrong ?

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

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

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

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.

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

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.