Guldin or it's not necessary? "surface" of revolution 3

Calculus Level 4

Consider the single rectangle in $\mathbb{R}^2$ that passes through the points $A = (1,2), B = (2,1), C = (4,3), D = (3,4)$ rotating around $x$ -axis in $\mathbb{R}^3$ .

The volume of the surface of revolution obtained can be written as $A\pi \text{ unit}^3$ .

Submit $A$ .

The answer is 20.

2 solutions

Nicola Mignoni
Feb 17, 2018

It's quite fast to evaluate the volume if we consider the situation in picture:

The volume can be calculated as the difference between the entire $AFCE$ shape and the $ABE$ and $CDF$ shapes, roteted around $x$ . So, it become

$\displaystyle V=\pi\left[ \int_{1}^{4} (x+1)^2-(x-1)^2 dx - \int_{1}^{2} (3-x)^2-(x-1)^2 dx-\int_{3}^{4} (x+1)^2-(7-x)^2 dx\right]=20\pi$

Otto Bretscher
Apr 4, 2016

Thanks to the genius of the great Swiss mathematician Habakkuk Guldin, we know that the volume is the area of the rectangle (which is 4) multiplied by the distance traveled by the centroid (which is $2\pi\times 2.5=5\pi$ ). The answer is $\boxed{20}$ .

Yes. that is, thank you (+1) $\uparrow$ . Or,haha, you can sum and substrate volume of "frustums" or "truncated cones".... I have done it in the two ways.

Guillermo Templado - 5 years, 2 months ago

Guldin or it's not necessary? "surface" of revolution 3

The answer is 20.

2 solutions

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