1. ## Surface area integral

Find surface area of solid formed by rotating the following curve about the y-axis:

$x = \sqrt{a^2 - y^2}$ on $0 \leq y \leq \frac{a}{2}$

*The "a" in there kinda screwed me up, but this is my work:

When I solved for the arc length, I ended up getting $\sqrt{2}$, like this:

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

$x' = \frac{2a - 2y}{2\sqrt{a^2-y^2}}$

$= \frac{a-y}{a-y} = 1$

$ds = \sqrt{1 + (1)^2}$

**Is this right?

So my integral is:

$\int\limits^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^2 - y^2}) (\sqrt{2})$

Then I pulled my square root out in front:

$2\sqrt{2} \pi \int\limits^{\frac{a}{2}}_{0} \sqrt{a^2 - y^2} dy$

$2\sqrt{2} \pi \int\limits^{\frac{\pi}{6}}_{0} asin \theta \cdot acos \theta d\theta$

I have more, but can some one check this up until this point?
Thanks!! Molly

2. Originally Posted by mollymcf2009
Find surface area of solid formed by rotating the following curve about the y-axis:

$x = \sqrt{a^2 - y^2}$ on $0 \leq y \leq \frac{a}{2}$

*The "a" in there kinda screwed me up, but this is my work:

When I solved for the arc length, I ended up getting $\sqrt{2}$, like this:

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

$x' = \frac{2a - 2y}{2\sqrt{a^2-y^2}}$

$= \frac{a-y}{a-y} = 1$ ****

$ds = \sqrt{1 + (1)^2}$

**Is this right?

So my integral is:

$\int\limits^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^2 - y^2}) (\sqrt{2})$

Then I pulled my square root out in front:

$2\sqrt{2} \pi \int\limits^{\frac{a}{2}}_{0} \sqrt{a^2 - y^2} dy$

$2\sqrt{2} \pi \int\limits^{\frac{\pi}{6}}_{0} asin \theta \cdot acos \theta d\theta$

I have more, but can some one check this up until this point?
Thanks!! Molly
From above **** $x' = \frac{2a - 2y}{2\sqrt{a^2-y^2}} = \frac{a - y}{\sqrt{a^2-y^2}} \ne \frac{a-y}{a-y}$.