Can any one help me with the following questions?
Use shell method to express the volume of the solid obtained by rotating the region bounded by the curves $x = \sqrt{\sin(8y)}$, $0<=y<=\pi/4$, and $x = 0$ about the line $y = 2$. (Don't need to compute the Integral.)
Use shell method to express the volume of the solid obtained by rotating the region bounded by the curves $x = \sqrt{\cos(8y)}$, $-\pi/8<=y<=\pi/8$, and $x = 0$ about the line $x = 4$.
(Don't need to compute the Integral.)
I tried to sketch the graph but it looks weird. Can someone help me and explain it to me? Thanks.
First of all, your $y$-range in both plots is way to large. Each integral should have only one loop. Now, for the sine case, the shell method must be integrate along the $y$-axis, relative to $y=2$, so in a formal sense
$$V=2\pi\int_0^2(y-2)f(x)dy$$
In other words, the shell element at a distance (2-y) has a volume of $2\pi (2-y)f(x)dx$. Then we can say
$$V=2\pi\int_0^2(y-2)\sqrt{\sin(8y)}dy=2\pi\int_0^{\pi/8}(2-y)\sqrt{\sin(8y)}dy$$
Notice the change in the integration limits since $f(x)$ is zero everywhere else.
For the cosine case, we write the formal shell integral as
$$V=2\pi\int_0^4(4-x)f(y)dx$$
Now, we have to be careful here because $f(y)$ is double-valued. Since it's symmetric we can take twice the the integral of the positive branch, then, since $y(x)=\cos^{-1}(x^2)/8$, we can write
$$V=(\pi/2)\int_0^4(4-x)\cos^{-1}x^2dx=(\pi/2)\int_0^1(4-x)\cos^{-1}(x^2) dx$$
where again we have a change in the integration limits because $f(y)=0$ for $x>1$.
I have verified these results by comparison with Pappus's $(2^{nd})$ Centroid Theorem.