area

**shilz222** · November 6th 2007, 09:04 AM

Find the area of the cone $z^2 = a^2(x^2 + y^2)$ between the planes $z = 1$ and $z = 2$ .

I think you use cylindrical coordinates.

So: $z^{2} = a^{2} r^{2}$

And is it: $\int_{0}^{2 \pi} \int_{0}^{\frac{z^{2}}{a^2}} \int_{1}^{2} a^{2}r^{2} \ dz \ dr \ d \theta$ ?

**Jhevon** · November 6th 2007, 09:17 AM

Originally Posted by shilz222

Find the area of the cone $z^2 = a^2(x^2 + y^2)$ between the planes $z = 1$ and $z = 2$ .

I think you use cylindrical coordinates.

So: $z^{2} = a^{2} r^{2}$

And is it: $\int_{0}^{2 \pi} \int_{0}^{\frac{z^{2}}{a^2}} \int_{1}^{2} a^{2}r^{2} \ dz \ dr \ d \theta$ ?

no. this is not a triple integral question, and you want the surface area.

use: $\mbox{Area} = \iint_D \sqrt{1 + \left( \frac {\partial z}{\partial x} \right)^2 + \left( \frac {\partial z}{\partial y} \right)^2}~dA$

D is the region we're integrating over. use polar coordinates

**shilz222** · November 6th 2007, 02:50 PM

So you let $z = \sqrt{a^{2}x^{2} + a^{2}y^{2}}$ and then:

$f_x = \frac{a^{2}x^{2}}{2}\left (a^{2}x^{2} + a^{2}y^{2}\right)^{-\frac{1}{2}}$

$f_y = \frac{a^{2}y^{2}}{2}\left(a^{2}x^{2} + a^{2}y^{2}\right )^{-\frac{1}{2}}$

So we have: $\int_{0}^{2 \pi} \int_{1/a}^{2/a} \sqrt{1 + \frac{a^{4}x^{4}}{4}(a^{2}x^{2} + a^{2}y^{2})^{-1}+ \frac{a^{4}y^{4}}{4}(a^{2}x^{2} + a^{2}y^{2})^{-1}}$

Then convert the inside into polar coordinates (which is a bit messy). Is this ok?

**shilz222** · November 7th 2007, 02:42 AM

Actually i ended up getting $\int_{0}^{2 \pi} \int_{1/a}^{2/a} \sqrt{1+a} \ r \ dr \ d \theta = \int_{0}^{2 \pi} d \theta \int_{1/a}^{2/a} r \sqrt{1+a^{2}} \ dr$

This equals $\frac{3 \pi \sqrt{1+a^{2}}}{a^{2}}$

Is this correct?

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