Calculus Calculus Math Forum

 November 4th, 2017, 05:22 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 surface integral problem Find the area of the cylinder $\displaystyle x^2 + z^2 = a^2$ that lies inside the cylinder $\displaystyle x^2 + y^2 = a^2$. Official Answer: $\displaystyle 8a^2$ My solution $\displaystyle x = r cos \theta , y = r sin \theta , z = \sqrt{a^2 - r^2cos^2 \theta }$ $\displaystyle \int_{0}^{2 \pi} \int_{0}^{a} \sqrt { \frac {(r^2cos \theta sin^2 \theta + r^2 cos^3 \theta)^2}{a^2 - r^2cos^2 \theta} + r^2} ~dr ~d \theta$ Is my solution correct? November 7th, 2017, 03:51 PM #2 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 The above integrand is incorrect. I think it should have been the following: $\displaystyle \int_{0}^{2 \pi} \int_{0}^{a} \sqrt{ \frac{( r^2 cos^3 \theta - r^2 cos \theta sin^s \theta )^2+(2r^2 cos^2 \theta sin \theta )^2 }{ a^2 - r^2 cos^2 \theta } + r^2 } ~~~dr ~d \theta$ Tags integral, problem, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zollen Calculus 9 September 5th, 2017 04:17 PM zollen Calculus 6 September 4th, 2017 01:47 PM zollen Calculus 2 September 4th, 2017 09:05 AM henrymerrild Calculus 2 May 1st, 2014 10:33 AM kriko Calculus 1 August 21st, 2010 10:55 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      