Alternate method using multivar also displayed
Let R be the region in the first quadrant bounded by the line $y=2x$, the curve $y=sin(x)$, and the line $l$ that is perpendicular to $y=2x$ and goes through the point ($\frac{\pi}2{}$,$1$). Find the volume of the solid $S$ obtained when $R$ is rotated $360$ degrees around the line $y=2x$. Also, find the surface area of $S$ including the circular base swept out by $l$.
I tried to use Pappus' Theorem to find the volume using the volume is equal to the area I found the area setting up two different integrals finding the area to equal $0.60217$. I then found the centroid coordinates at $(.8409, .7756)$ and the distance to equal 0.164 and the circumference the centroid travels to then equal $1.03195$ and the only part I am sure of is the arc length is $1.9101$.
Obviously $1.03195$ x $0.60217$ $\neq$ $2.74$ and $1.03195$ x $1.9101$ $\neq$ $7.5$ that wolfram alpha found. Can someone help me realize my mistake? I think the error is the centroid location or centroid distance traveled. A centroid radius of around $0.62$ makes the math work out perfectly.
The Wolfram Alpha solution:
[Image of the question in detail to eliminate confusion][2]
i.stack.imgur.com/Q93OA.png
I don't have enough reputation just add an https:// before the i.stack
Half a spinning top!
To use Pappus' theorem, you need to find the coordinates of several points in this problem. As describing the computation of all that can be quite verbose, here is a figure to set things up and am providing the coordinates I got - we can discuss if you got something different
length of sine curve:
$$S=\int_0^{\frac{\pi}{2}}\sqrt{1+\cos^2x}dx\\
x_L=\frac{1}{S}\int_0^{\frac{\pi}{2}}x\sqrt{1+\cos^2x}dx\\
y_L=\frac{1}{S}\int_0^{\frac{\pi}{2}} \sin x\sqrt{1+\cos^2x}dx
$$
Area to be rotated:
$$A=\int_0^{X_M} (2x-\sin x) dx + \int_{x_M}^\frac{\pi}{2}(\frac{-x}{2}+1+\frac{\pi}{4}-\sin x) dx\\
=0.2657+0.2845=0.5502
$$
Centroid:
$$A \times x_C=\int_0^{X_M} x(2x-\sin x) dx + \int_{x_M}^\frac{\pi}{2}x (\frac{-x}{2}+1+\frac{\pi}{4}-\sin x) dx\\
2A \times y_C=\int_0^{X_M} (2x)^2-\sin^2 x) dx + \int_{x_M}^\frac{\pi}{2} \left[(\frac{-x}{2}+1+\frac{\pi}{4})^2-\sin^2 x \right] dx
$$
Results:
$$M=(x_m,y_m)=\left(\frac{4+\pi}{10},\frac{4+\pi}{5}\right)\\
R=(\frac{\pi}{2},1)\\
L=(0.7317,0.6009)\\
C=(0.7359,0.8870)\\
d_C=0.2615\\
d_L=0.3857\\
d_R=0.9577\\
A=0.5501\\
S=1.9101
$$
where $A$ is the total area (pink+green) and $S$ is the length of the $\sin$ curve from the origin to R. From here, the volume is $$V=2\pi d_CA=0.9041\\ SA=2\pi d_L S + \pi d_R^2=7.5106$$
Note: This problem is straightforward in principle. Most of the work here is only because of the tilted axis of rotation. I usually avoid using the Pappus rule and prefer to do integration directly. However, in this case, it seems that it's not easy to get a closed form (impossible so far for me) for the sine curve if you do a coordinate transformation to make the axis of rotation the vertical axis of the coordinate system. Hence, we need to resort to Pappus.