Teardrop

Calculus Level 4

$\large f(x) = \sqrt x - \tfrac12x.$ A teardrop-shaped solid is obtained by taking the part of the graph of $f$ that lies above the $x$ -axis and revolving it around that axis.

Calculate the volume of this solid.

The answer is 1.6755.

1 solution

Arjen Vreugdenhil
Feb 16, 2016

First, we find the boundaries for the integration. The left boundary is clearly $a = 0$ ; the right boundary is the other zero of the function, which is $b = 4$ , because $\sqrt 4 - \tfrac12 \cdot 4 = 2 - 2 = 0$ .

$V = \pi \int_a^b \left(f(x)\right)^2\:dx \\ = \pi\int_0^4 \left(\sqrt x - \tfrac12x\right)^2\:dx \\ = \pi\int_0^4 (x - x\sqrt x + \tfrac14x^2)\:dx \\ = \pi\left[\frac12x^2 - \frac25x^2\sqrt x + \frac1{12}x^3\right]_0^4 \\ = \pi\left(8 - \frac{64}5 + \frac{16}3\right) \\ = \frac{8\pi}{15} \approx \boxed{1.6755}.$

Or use Pappas's centroid volume theorem in which the volume is the distance the geometric centroid makes in one revolution times the area between the axis and the function being revolved.

Lance Rodewald - 5 years, 3 months ago

That is an interesting theorem, but does it help? Calculating the $y$ -coordinate of the centroid is at least as much work, because one would have to integrate $f(x)$ itself as well: $\langle y\rangle = \frac{\int_a^b \tfrac12(f(x))^2\:dx}{\int_a^b f(x)\:dx} = \dots$ Finding the distance traveled by the centroid removes the factor $\tfrac12$ and adds a factor $\pi$ ; multiplying by the area removes the denominator, and we are back to the original integral!

Arjen Vreugdenhil - 5 years, 3 months ago

Teardrop

The answer is 1.6755.

1 solution

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