R racewithferrari Nov 2009 92 1 Jul 11, 2010 #1 the endpoint comes -1 and 1 and I am taking the integral of ((9/8)-x^2)-(x^2/8)....and then multiplyig by 2pi. the final answers come 3pi. Any help thanks

the endpoint comes -1 and 1 and I am taking the integral of ((9/8)-x^2)-(x^2/8)....and then multiplyig by 2pi. the final answers come 3pi. Any help thanks

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Jul 11, 2010 #2 method of washers ... \(\displaystyle \displaystyle V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx\) Reactions: racewithferrari

R racewithferrari Nov 2009 92 1 Jul 11, 2010 #3 skeeter said: method of washers ... \(\displaystyle \displaystyle V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx\) Click to expand... so its like this with end points of -1 and 1

skeeter said: method of washers ... \(\displaystyle \displaystyle V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx\) Click to expand... so its like this with end points of -1 and 1

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Jul 11, 2010 #4 racewithferrari said: so its like this with end points of -1 and 1 Click to expand... second term in the integrand should be \(\displaystyle \left(\frac{x^2}{8}\right)^2\) if what you posted originally is correct ... \(\displaystyle \frac{x^2}{8} \ne \frac{1}{8x^2}\) also ... limits of integration? Reactions: racewithferrari

racewithferrari said: so its like this with end points of -1 and 1 Click to expand... second term in the integrand should be \(\displaystyle \left(\frac{x^2}{8}\right)^2\) if what you posted originally is correct ... \(\displaystyle \frac{x^2}{8} \ne \frac{1}{8x^2}\) also ... limits of integration?