# Math Help - Calculus Question

1. ## Calculus Question

If anyone could explain how this problem is done, it would be very much appreciated!

Write the equation for the surface generated by revolving the given around the indicated axis.

4x + 9y^2 = 36 around the y-axis

2. Originally Posted by meebo0129
If anyone could explain how this problem is done, it would be very much appreciated!

Write the equation for the surface generated by revolving the given around the indicated axis.

4x + 9y^2 = 36 around the y-axis
You want the equation of the surface or the surface area?

-Dan

3. The equation of the surface

4. Originally Posted by meebo0129
If anyone could explain how this problem is done, it would be very much appreciated!

Write the equation for the surface generated by revolving the given around the indicated axis.

4x + 9y^2 = 36 around the y-axis
The solid (I can't think of a name for this thing) has, as a cross-section on the xy plane a pair of parabolas both with vertices at the origin and axes of symmetry along the +x and -x axes. As we move along the y direction the cross-sections in the xz plane will be circular, so we need the equation of a circle in the xz plane centered on the origin. That would be
$x^2 + z^2 = r^2$
where r is the radius of the circle in the cross-section, which will depend on the value of y we choose for the cross-section. The radius will be equal to the distance from the point on the surface to the y-axis, which in this case will be measured by the x value. Thus the radius is given by the original equation:
$r = x = -\frac{9}{4}y^2 + 9$
and the surface becomes:
$x^2 + z^2 = \left ( -\frac{9}{4}y^2 + 9 \right )^2$

or
$x^2 - \left ( \frac{9}{4}y^2 - 9 \right )^2 + z^2 = 0$

-Dan