Dec 16, 2013 Thread starter #1 J jk8985 New member Dec 16, 2013 12 Let E be the solid inside cylinder y^2+z^2=1 and x^2+z^2=1, find the volume of e and the surface area of e
Let E be the solid inside cylinder y^2+z^2=1 and x^2+z^2=1, find the volume of e and the surface area of e
Dec 16, 2013 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,775 Can you show us what you have tried so our helpers know where you are stuck?
Dec 16, 2013 Thread starter #3 J jk8985 New member Dec 16, 2013 12 For the second half of the question, I got this. Is it correct?? Also, could you write it out nicely I don't really understand what I did. Can someone let me know if this is correct and if I showed all steps? Since the the region of integration inside x^2 + y^2 = 1 (and symmetry), convert to polar coordinates: 2 * 4 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) (r dr dθ)/(1 - r^2 cos^2(θ))^(1/2) = 8 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) 2r (1 - r^2 cos^2(θ))^(-1/2) dr dθ = 8 * ∫(θ = 0 to π/2) (-1/cos^2(θ)) 2(1 - r^2 cos^2(θ))^(1/2) {for r = 0 to 1} dθ = 16 ∫(θ = 0 to π/2) (1/cos^2(θ)) [1 - (1 - cos^2(θ))^(1/2)] dθ = 16 ∫(θ = 0 to π/2) (1 - sin θ) dθ/cos^2(θ) = 16 ∫(θ = 0 to π/2) (sec^2(θ) - sec θ tan θ) dθ = 16 (tan θ - sec θ) {for θ = 0 to π/2} = 16 (sin θ - 1)/cos θ {for θ = 0 to π/2} = 16 (0 - (-1)), using L'Hopital's Rule as θ→ π/2- = 16.
For the second half of the question, I got this. Is it correct?? Also, could you write it out nicely I don't really understand what I did. Can someone let me know if this is correct and if I showed all steps? Since the the region of integration inside x^2 + y^2 = 1 (and symmetry), convert to polar coordinates: 2 * 4 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) (r dr dθ)/(1 - r^2 cos^2(θ))^(1/2) = 8 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) 2r (1 - r^2 cos^2(θ))^(-1/2) dr dθ = 8 * ∫(θ = 0 to π/2) (-1/cos^2(θ)) 2(1 - r^2 cos^2(θ))^(1/2) {for r = 0 to 1} dθ = 16 ∫(θ = 0 to π/2) (1/cos^2(θ)) [1 - (1 - cos^2(θ))^(1/2)] dθ = 16 ∫(θ = 0 to π/2) (1 - sin θ) dθ/cos^2(θ) = 16 ∫(θ = 0 to π/2) (sec^2(θ) - sec θ tan θ) dθ = 16 (tan θ - sec θ) {for θ = 0 to π/2} = 16 (sin θ - 1)/cos θ {for θ = 0 to π/2} = 16 (0 - (-1)), using L'Hopital's Rule as θ→ π/2- = 16.