I know this is a lot of stuff, but I want to be certain that I have all of this information correct. If there are any errors, I feel like they would come from the first problem or from $\displaystyle \frac{d^2y}{dx^2}$ on the last one. Many thanks to anyone willing to help me out with any of this.

1)Find the surface area of the solid obtained by revolving $\displaystyle \frac{1}{4}x = \sqrt{1-y}$ for $\displaystyle -1\leq y \leq 0$ about the y-axis

$\displaystyle x = 4\sqrt{1-y}$

$\displaystyle \frac{dx}{dy} = \frac{2}{\sqrt{1-y}}$

Surface Area $\displaystyle = 2\pi\int_{-1}^{0}4\sqrt{1-y}\left(\frac{2}{\sqrt{1-y}}\right)^2dy$

$\displaystyle = 8\pi\int_{-1}^{0}\sqrt{1-y}\left(\frac{4}{1-y}\right)dy$

$\displaystyle = 32\pi\int_{-1}^{0}\frac{\sqrt{1-y}}{1-y} dy$

$\displaystyle = 32\pi\int_{-1}^{0}\frac{1}{\sqrt{1-y}}dy $

$\displaystyle = -32\pi\left.(2\sqrt{1-y})\right|_{-1}^{0}$

$\displaystyle = -64\pi\left.(\sqrt{1-y})\right|_{-1}^{0}$

$\displaystyle = -64\pi-(-64\pi\sqrt{2})$

$\displaystyle = 64\pi(\sqrt{2}-1)$

2)Find an equation relating $\displaystyle x$ and $\displaystyle y$. Then sketch the curve C whose parametric equations are given and indicate the direction as $\displaystyle t$ increases

a)$\displaystyle x = 2t-1$, $\displaystyle y=8t^3-1$ for all t

$\displaystyle t=\frac{x+1}{2}$

$\displaystyle y=8\left(\frac{x+1}{2}\right)^3-1$

$\displaystyle y=(x+1)^3-1$

$\displaystyle y=x^3+3x^2+3x+1-1$

$\displaystyle y=x^3+3x^2+3x$

My graph looks like an elongated $\displaystyle y=x^3$ graph which goes through (0,0) and has a horizontal tangent at (-1,-1). The direction of t is going upwards.

b)$\displaystyle x=sin^2t,$ $\displaystyle y=cos t$ for $\displaystyle 0 \leq t \leq \pi$

$\displaystyle x + y^2 = 1$

$\displaystyle y = \sqrt{1-x}$ not $\displaystyle \pm$ since $\displaystyle 0 \leq x \leq 1$

My graph for this one is the right half of a circle of radius 1 centered on the origin. The direction of t is going clockwise.

3)Let C be the curve with parametrization $\displaystyle x=12t-t^3,$ $\displaystyle y=t^2-4$ for all t

Find equations of the tangent lines to C at the points corresponding to $\displaystyle t=1$ and $\displaystyle t=2$. At what point(s) will the graph have a horizontal tangent line? Also find $\displaystyle \frac{d^2y}{dx^2} $. Use all this information to sketch the curve C.

$\displaystyle \frac{dy}{dx}= \frac{2t}{12-3t^2}$

for $\displaystyle t=1$, $\displaystyle m=\frac{2}{9}$ and equation of the tangent line is $\displaystyle y+3=\frac{2}{9}(x-11)$

for $\displaystyle t=2$, m is undefined and the equation of the tangent line is $\displaystyle x=-16$

$\displaystyle \frac{d^2y}{dx^2} = \frac{\frac{d}{dx}[2t(12-3t^2)^{-1}]}{12-3t^2} $

$\displaystyle = \frac{(2t)(-1)(12-3t^2)^{-2}(-6t)+2(12-3t^2)^{-1}}{12-3t^2}$

$\displaystyle = \frac{2(12-3t^2)^{-1}[1+6t^2(12-3t^2)^{-1}]}{12-3t^2}$

$\displaystyle = \frac{2[1+6t^2(12-3t^2)^{-1}]}{(12-3t^2)^2}$

$\displaystyle = \frac{2[1+\frac{6t^2}{12-3t^2}]}{(12-3t^2)^2}$

For the graph I have vertical tangents and x intercepts at $\displaystyle t=\pm2$ (-16, 0), (16, 0), a horizontal tangent at $\displaystyle t=0$ (0, -4). The graph crosses itself at the y-intercept (0, 8), is concave down while above the x-axis and concave up when below. It makes a big loop from left to right.

Any help is enormously appreciated (Wink)(Clapping)