MHB Double Integral Problem/ Surface Area of parametric surface

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The discussion focuses on solving a double integral related to the surface area of a parametric surface. The user has successfully computed the cross product but is struggling with the double integral. A response clarifies that the inner integral can be evaluated first, simplifying the process. The final expression for the integral is presented as π times the integral of the square root of (1 + u²) from 0 to 1. This highlights a method for tackling double integrals in surface area calculations.
Dizzy1
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Hi I'm doing a surface area problem with a parametric surface and I got the cross product but I can't figure out the double integral.

I found the solution online but with no explanation, so can someone explain how to solve this integral:

View attachment 2449

thank you!
 

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Dizzy said:
Hi I'm doing a surface area problem with a parametric surface and I got the cross product but I can't figure out the double integral.

I found the solution online but with no explanation, so can someone explain how to solve this integral:

View attachment 2449

thank you!

Because the inner integral is just a number you can immediately do the outer integral so:

$$\int_0^{\pi}\ \int_0^1\sqrt{1+u^2}du\ dv=\pi \int_0^1\sqrt{1+u^2}du$$

.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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