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.
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
Because r = cos(theta) and when theta = 0, we are at the point (1,0) and when theta is pi/2, we are at the point (0,0). Therefore the semicircle is traced from r = 0 to r = pi/2.
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).
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]
this guy is awesome. he goes at the perfect pace. some videos out there are way too slow.
Many thanks mate... you are surely preparing me for my exam with your thorough steps :)
Your videos are saving my Calc III class. I will definitely will recommend it to my friends!
+Julio C. S Glad I could help. Thank you for the recommendations.
great video , I wonder what is the name of the programe to get this calculator you have used through the video ?
Thanks for your videos ,very helpful as usual and btw i like the quotes at the end, really inspiring. God bless you.
U just saved my grades there mate....thnx😊
Why does the limit of theta vary from 0 to ¥/2? Shouldn't it be from 0 to ¥? (¥=pi)
My question exactly....
Theta is measured from origin, not from center of circle.
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.
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
why isn't the limit of r not 0 to 1/2 considering that 1/2 is the radius of the region?
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
Because r = cos(theta) and when theta = 0, we are at the point (1,0) and when theta is pi/2, we are at the point (0,0). Therefore the semicircle is traced from r = 0 to r = pi/2.
But how do we know that we are traveling from the point (1,0) to (0,0) if we are not allowed graphing calculators on the test?
why we use r=0 to r=cos(theta) instead of r=one half ? it is just a semicircle
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.
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 :)
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).
Thanks a lot. Keep the videos going!
this is better than patrickJMT and khan
the point is the sketch is in the first quadrant which is 90 degrees..so it'll back at its origin :)
Love your videos!!!
Your are intelligence man.
Thanks for your time sr!!!
i wish you were my teacher.
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
You are amazing..
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]
@rmwhite13691 Same problem here
Thank You Very Much!