R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Jul 21, 2010 #1 i posted this problem earlier but deleted it. i did it again and i am confident that this time its right. what do you think. Attachments lastscan.jpg 91.3 KB · Views: 44
i posted this problem earlier but deleted it. i did it again and i am confident that this time its right. what do you think.
R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Jul 21, 2010 #2 i just turned in my quiz and got the above problem correct. and scored a 100 on my quiz.
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Jul 21, 2010 #3 \(\displaystyle Disc: \ Revolved \ around \ x \ = \ 0 \ (y-axis).\) \(\displaystyle \pi\int_{0}^{1}[((y+9)/10)^2-(y^2)^2]dy+\pi\int_{-9}^{0}[(y+9)/10]^2dy \ = \ \frac{47\pi}{15}\) \(\displaystyle Shell: \ Revolved \ around \ x \ = \ 0 \ (y-axis).\) \(\displaystyle 2\pi\int_{0}^{1}x[x^{1/2}-(10x-9)]dx \ = \ \frac{47\pi}{15}\)
\(\displaystyle Disc: \ Revolved \ around \ x \ = \ 0 \ (y-axis).\) \(\displaystyle \pi\int_{0}^{1}[((y+9)/10)^2-(y^2)^2]dy+\pi\int_{-9}^{0}[(y+9)/10]^2dy \ = \ \frac{47\pi}{15}\) \(\displaystyle Shell: \ Revolved \ around \ x \ = \ 0 \ (y-axis).\) \(\displaystyle 2\pi\int_{0}^{1}x[x^{1/2}-(10x-9)]dx \ = \ \frac{47\pi}{15}\)
