Find the equation for the surface of revolution by revolving

**icelated** · June 6th 2012, 11:49 AM

Find the equation for the surface of revolution by revolving
$2z = \sqrt {4- x^2}$ about the x- axis

Okay, this is far as i can get

$z = \frac {\sqrt {4- x^2}} {2}$

What should i do to solve this?

**Reckoner** · June 6th 2012, 12:28 PM

Originally Posted by icelated

Find the equation for the surface of revolution by revolving
$2z = \sqrt {4- x^2}$ about the y - axis

Are you sure that the question asks for revolution about the y-axis? Because that doesn't make a lot of sense to me. The equation $2z = \sqrt{4-x^2}$ defines a plane curve in the $xz$ -plane, which would be perpendicular to the $y$ -axis. The axis of revolution should be in the same plane as the generating curve.

**icelated** · June 6th 2012, 12:32 PM

Sorry, i meant x-axis

**Reckoner** · June 6th 2012, 12:47 PM

Originally Posted by icelated

Sorry, i meant x-axis

Okay, that's better.

Let's let $z = r(x) = \frac{\sqrt{4-x^2}}2$ . For a fixed point on this curve $\left(x_0, 0, r\left(x_0\right)\right)$ , revolving the point about the $x$ -axis produces a circle parallel to the $yz$ -plane whose equation is $y^2+z^2=\left[r(x_0)\right]^2$ . Replacing $x_0$ with the independent variable $x$ gives us an equation for the resulting surface of revolution:

$y^2+z^2=\left[r(x)\right]^2$

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

$\Rightarrow y^2 + z^2 = \frac{4-x^2}4$

$\Rightarrow \frac{x^2}4 + y^2 + z^2 = 1$

As you can see from the equation, this surface is an ellipsoid.

**icelated** · June 6th 2012, 01:30 PM

Thank you

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