Let A be the region in the first quadrant bounded by the curve $y=\cos x$, the line tangent to $y=\cos x$ when $x=\frac{\pi}{4}$ and the y-axis.
a) We now rotate A with respect to the line $y=-1$. Set up but do not evaluate the integral that represents the volume using cross sections.
b) We now rotate A with respect to the line $y=-1$. Set up but do not evalate the integral that represents the volume using cylindrical shells.
My work:
The line tangent to $y=\cos x$ is found by:
$y'=-\sin x$
$y'(\pi/4)=-\sin(\pi/4)=-\frac{\sqrt{2}}{2}$
$y-\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}(x-\frac{\pi}{4}$)
This is the equation of that line. The graph of this looks like:
PART A
For part a) is it asking to use the washer method?
The answer for part a) is
$$\pi\int_{0}^{\pi/4}(-\frac{\sqrt{2}}{2}(x-\pi/4)+\frac{\sqrt{2}}{2}+1)^2-(1+\cos x)^2 dx$$
I know that washers will be vertical and shells will be horizontal.
So for the washer method, the outer radius is
$R(y)=\cos x - (-1) = \cos x+1$
The inner radius is:
$r(y) = -\frac{\sqrt{2}}{2}(x-\pi/4)+1$
So the volume is given by
$\pi \int_{0}^{\pi/4} (\cos x+1)^2 - (-\frac{\sqrt{2}}{2}(x-\pi/4)+1)$
So I don't understand where I went wrong and why I get a different answer.
PART B
The answer is: $$\int_{a}^{b} 2\pi x(y) (y+1)dy+\int_{b}^{c} 2\pi(x(y)-\arccos y)(y+1)dy$$
where $a=\frac{\sqrt{2}}{2}, b=1, c=\frac{\sqrt{2}}{2}*\frac{\pi}{4}+\frac{\sqrt{2}}{2}$ and $x(y)=-\frac{2}{\sqrt{2}}(y-\frac{\sqrt{2}}{2})+\frac{\pi}{4}$
I'm really confused about this second part. I don't understand the bounds of integration. I know that since I'm using the shell method, everything needs to be in terms of $y$ so $x(y)$ is the inverse tangent line.
But I'm confused about how to set this integral up and where each part comes from.