Professor Organic Chemistry Tutor, thank you for a solid explanation and analysis of the Surface Area of Revolution of Polar Curves in Calculus Two. Reading and doing problems from a variety of Calculus textbooks, I cannot remember seeing this classic material on Surface Area of Revolution of Polar Curves. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
At 4:35 I'm not sure if I miss anything here but aren't the equations in line 2 and 3 wrong? Bacause the right side of the equations are not the value of the surface area. Also, the right side of the equation in line 3 and 4 are not equal. I don't think you can move r in and out of the integral like that since it is a function of zeta.
He moves the 10(sintheta) from the left to the right. He divides 10(sintheta) by 5 in order to make it 2(sintheta) to match the double angle identity utilized above. (Multiplies factor of 2pi by 5 to become 10pi) Since he initially regarded it a constant, the factor can be moved around inside and outside of the integral's bounds, but eventually it has to be integrated as sin is not a constant factor.
Next Video: th-cam.com/video/OPSCKXXvWiM/w-d-xo.html
Professor Organic Chemistry Tutor, thank you for a solid explanation and analysis of the Surface Area of Revolution of Polar Curves in Calculus Two. Reading and doing problems from a variety of Calculus textbooks, I cannot remember seeing this classic material on Surface Area of Revolution of Polar Curves. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
At 4:35 I'm not sure if I miss anything here but aren't the equations in line 2 and 3 wrong? Bacause the right side of the equations are not the value of the surface area.
Also, the right side of the equation in line 3 and 4 are not equal. I don't think you can move r in and out of the integral like that since it is a function of zeta.
Everything that Professor Organic Chemistry Tutor did Mathematically is legal. Please rewatch the video and takes notes.
At 4:17, how can you move 2 sin (theta) to the right side of the integral?
He moves the 10(sintheta) from the left to the right. He divides 10(sintheta) by 5 in order to make it 2(sintheta) to match the double angle identity utilized above. (Multiplies factor of 2pi by 5 to become 10pi)
Since he initially regarded it a constant, the factor can be moved around inside and outside of the integral's bounds, but eventually it has to be integrated as sin is not a constant factor.
What if the revolution occurs at theta=pi or theta=3pi/2, what formula do we use then, or do we tweak the formulas shown here somehow?
this was perfect, thanks!
What if they didnt give the upper and lower limit? How to determine their limit?
See where they intersect by setting the equations equal to each other. You can also graph it to see its x values
Thank you soo much.
And also you got any social media?
well does he
❤❤
U just factor out the 100
Sin aquare that's plus cos square that's equala to 1