Can someone plz help me out with inequality using polar coordinates?

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.

A community for students.

Can someone plz help me out with inequality using polar coordinates?

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

answer on brainly

SEE EXPERT ANSWER

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

1 Attachment
um... not really sure if I get the question, but the area swept out by r is\[\frac38\le r\le\frac12\]and theta is just\[0\le\theta\le2\pi\]
no sorry inner r is 3/16

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[\frac3{16}\le r\le\frac12\]\[0\le\theta\le2\pi\]I suppose
yes that is the answer
do you need explanation?
lol yes i dont understand how u got thise values
well polar coordinates are defined with radius and angle. Here are the diameters|dw:1329346620272:dw|so we know the radii are half this. therefor radius of any point on the washer is between 3/16inches (the radius of the inner circle) and 1/2inches (the radius of the outer circle)\[\frac3{16}\le r\le\frac12\]
ok got that
|dw:1329346791420:dw|as for the angle, the washer is circular, so the angle has to sweep out a complete circle:\[0\le\theta\le2\pi\](we never go around the circle more than once, so no need to put more than 2pi))
ok i think i get it
if it was top half of a washer, the angle would be \[0\le\theta\le\pi\]if it was a disk (no hole in the middle) the radius would be\[0\le r\le\frac12\]
ohhhh i see i get it
oh it is so simple ok Thanks for helping me out :)
anytime!

Not the answer you are looking for?

Search for more explanations.

Ask your own question