MHB Find the volume by using shell and disk method

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The discussion focuses on verifying the calculations for volume using both the shell and disk methods. For the shell method, the volume is calculated as V = 2π ∫ from 0 to 1 of (x+2)·x dx plus 2π ∫ from 1 to 4 of (x+2)·√x dx. For the disk method, the volume is expressed as V = π ∫ from 0 to 1 of (6^2 - (y+2)^2) dy plus π ∫ from 1 to 2 of (6^2 - (y^2+2)^2) dy. The user seeks confirmation on the accuracy of these calculations. The discussion emphasizes the importance of correct application of integration techniques in volume calculations.
jaychay
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Can you check it for me that I done it right or not ?
Thank you in advance.
 
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shells ...

$\displaystyle V = 2\pi \int_0^1 (x+2) \cdot x \, dx + 2\pi \int_1^4 (x+2) \cdot \sqrt{x} \, dx$

washers ...

$\displaystyle V = \pi \int_0^1 6^2 - (y+2)^2 \, dy + \pi \int_1^2 6^2 - (y^2+2)^2 \,dy$
 

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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