Volume of solids of revolution (1 Viewer)

RoXaH · Jun 10, 2009

The region bounded by the curve $x^{\frac{1}{2}} +y^{\frac{1}{2}} = 2$ and the coordinate axes is rotated about the X-axis. Find the volume of the solid generated. Answer is 64pi/15

kurt.physics · Jun 10, 2009

RoXaH said:
The region bounded by the curve $x^{\frac{1}{2}} +y^{\frac{1}{2}} = 2$ and the coordinate axes is rotated about the X-axis. Find the volume of the solid generated. Answer is 64pi/15

We have:

$\sqrt{x} + \sqrt{y} = 2$

Note that:

$x \geq 0\,\, \forall x$ and $y \geq 0\,\, \forall y$

Futhermore,

$\sqrt{y} = 2 - \sqrt{x}$

$\left(\sqrt{y}\right)^2 = \left( 2 - \sqrt{x} \right)^2$

$\therefore y = 4 - 4\sqrt{x} + x$

The x-intersept occurs when y = 0, hence

$\sqrt{x} + \sqrt{0} = 2$

$\sqrt{x} = 2$

$\therefore x = 4$

(Also the y-intercept is 4)

Hence to find the volume generated all we have to do is rotate the function about the x-axis between 0 and 4.

$y^2 = \left(4 - 4\sqrt{x} + x\right)^2$

$= \left(4 - 4\sqrt{x} + x\right)\left(4 - 4\sqrt{x} + x\right)$

$= 16 - 16\sqrt{x} + 4x - 16\sqrt{x} + 16x -4x\sqrt{x} + 4x -4x\sqrt{x} + x^2$

$=x^2 + 24x -8x\sqrt{x} - 32\sqrt{x} + 16$

$=x^2 + 24x - 8x^{\frac{3}{2}} - 32x^{\frac{1}{2}} + 16$

$V = \pi \int_{a}^{b} y^2 dx$

$= \pi \int_{0}^{4} \left(x^2 + 24x - 8x^{\frac{3}{2}} - 32x^{\frac{1}{2}} + 16\right) dx$

$= \pi \cdot \left[ \frac{x^3}{3} + 12x^2 - \frac{16x^\frac{5}{2}}{5} - 32 \cdot \frac{2x^\frac{3}{2}}{3} + 16x\right]_{0}^{4}$

$= \pi \cdot \left[ \frac{4^3}{3} + 12\cdot 4^2 - \frac{16\cdot 4^\frac{5}{2}}{5} - 32 \cdot \frac{2\cdot 4^\frac{3}{2}}{3} + 16\cdot 4 \right]$

(i excluded the subtraction of the part where you substitute zero because you can easily see that the result is zero)

$= \pi \cdot \frac{64}{15}$

$= \frac{64\pi}{15}$

$\approx 13.4$

Ostentatious · Jun 12, 2009

Whats with the upside down A?

gurmies · Jun 12, 2009

It means "for all values of"

Ostentatious · Jul 20, 2009

Ah, thanks for that. I usually use $\varepsilon R$ .

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Volume of solids of revolution (1 Viewer)

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