Double Integrals in Polar Coordinates - Example 2

This is a second example of how to rewrite a double integral given in rectangular form in polar form.

[Yoaedn]

this guy is awesome. he goes at the perfect pace. some videos out there are way too slow.

[djjcyxz]

Your videos are saving my Calc III class. I will definitely will recommend it to my friends!

[lfcforever]

Many thanks mate... you are surely preparing me for my exam with your thorough steps :)

[safaaaz]

Thanks for your videos, very helpful as usual and btw i like the quotes at the end, really inspiring. God bless you.

[kushagrasharma]

U just saved my grades there mate....thnx😊

[CocoGras]

Maybe I'm missing something, but 1/24 seems kind of small for the area of half of a circle with radius of 1/2. If area of a circle is pi*radius <squared, then wouldn't the area of this particular half circle be ~3/8?

[Name-pnrf]

Why does the limit of theta vary from 0 to ¥/2? Shouldn't it be from 0 to ¥? (¥=pi)

[fawzyhegab]

great video, I wonder what is the name of the programe to get this calculator you have used through the video ?

[darcash]

we see that x is from 0 to 1, placing it somewhere in the first or third quadrants. and that y has a sqrt at the top, placing it in the first or second. Combining these facts, it is solely in the first.
thus we get theta bounds: [0, pi/2].

Then we see that y has lower bound 0, upper bound sqrt(x-x^2).
y = 0
y = sqrt(x-x^2)
--> sqrt(x-x^2) = 0
x - x^2 = 0
r(costheta - r) = 0
--> r = 0, r = costheta.
these are the bounds for our r: [0, costheta]

[rmwhite]

I keep getting burned by these types of problems where the lower bound of integration for the internal integral is the line y = x. How do I convert this equation and solve for r? The only answer I come up with is r = r*tan(theta), which just turns out to be garbage since you're now trying to evaluate the partial integral by replacing an r with another r.

[piemaster]

from 4:08-5:13 I just used the previous formula you had x^2+y^2=x to get r^2=rcostheta. then it works the same :)

[terekingli]

why isn't the limit of r not 0 to 1/2 considering that 1/2 is the radius of the region?

[SN-thbo]

Mathispower4u Thank you for this video- just wondering why theta is equal to 90 or pi/2 and not pi? When I graph it in rectangular just like the previous example we were able to get theta from the graph. Thank you

[znhait]

Well, y = x passes through the angle pi/4. if it was a circle centered at the origin r would go from 0 to whatever the radius is.

[johnvalgon]

why we use r=0 to r=cos(theta) instead of r=one half ? it is just a semicircle

[eddywangchang]

this is better than patrickJMT and khan

[NeverLieToYa]

the point is the sketch is in the first quadrant which is 90 degrees..so it'll back at its origin :)

[sabrinacamargo]

I found it easier to replace the function that's being integrated with the square root of r^2 rather than plug in all that.
r^2 = x^2 + y^2

[Nowhere]

at 6:04, '' and when theta is 90 degrees or pie/2 we are back at the origin'' ???i
when theta is 180 degrees or pie we are back at the origin.
whe theta is 90 degrees we are at ( 1/2, 1/2).