Recent content by Spout

it turned out correct, you guys are legends

I see well I am guessing 0 ≤ r ≤ 2 because this covers the min and max radius and 0 ≤ θ ≤ 2pi because this covers all points angular-ly then the integral would be ∫(0 to 2pi) ∫(0 to 2) [ √( 1 + 4r²cos²θ + 4r²sin²θ) ] rdrdθ ∫(0 to 2pi) ∫(0 to 2) [ √( 1 + 4r²) ] rdrdθ :)? amirite

This is quite embrassing but what does "Since you are projecting onto the x-y plane, find zx and zy hence find the surface area" mean up to now i only know the process of solving these problems i don't know what partial derivative represents or means, so I'm unable to comprehend this sentence...

Homework Statement I need to evaulate ∫ ∫S dS where S is the surface z = x² + y², 0 ≤ z ≤ 4. Homework Equations dS = √( 1 + ƒ²x + ƒ²y)dxdyThe Attempt at a Solution dS = √( 1 + 4x² + 4y²)dxdy here's the problem what are the limits to the surface integral? no clue.. dx means i should find...