Revolve $$ x^2 + 4y^2 = 4$$
a. About y = 2.
b. About x = 2.
The answer at the back of the book is the same for both
$$
a. 2\pi \int_{0}^{2} 8\sqrt{1-\frac{x^{2}}{4}} dx = 78.95684
$$
$$
b. 2\pi \int_{0}^{1} 8\sqrt{4-4{y^{2}}} dy = 78.95684
$$
I tried splitting the ellipse into its upper and lower half but I still couldn't get the answer. I get that its symmetrical that's why it's multiplied to two but I don't know how you get the equation inside the integral. Thanks in advance.
On
Let's try the shell method instead of the disk method? Let's consider the problem b).
The distance from the axis of rotation is (2-x). Considering the top half of the ellipse, $$ A = 2\pi\int^2_{-2} (2-x)\sqrt{1-\frac{x^2}{4}} \ \mathrm{d}x = 4\pi^2 $$ Doubling this gives us the volume of the whole ellipsoid, $8\pi^2 \approx 78.95$
I will think about the disk method and maybe add it on.
About $y=2$: It seems natural to take cross-sections perpendicular to the $x$-axis. A cross-section at $x$ is a washer, with outer radius $R(x)=2+\sqrt{1-\frac{x^2}{4}}$ and inner radius $r(x)=2-\sqrt{1-\frac{x^2}{4}}$.
The volume is $$\int_{-2}^2 (\pi R^2(x)-\pi r^2(x))\,dx.$$ When we expand and simplify, there is a lot of cancellation, just like in the first problem, and we get $$\int_{-2}^2 8\pi \sqrt{1-\frac{x^2}{4}}\,dx.$$ It is a good idea to take advantage of symmetry. So we integrate from $0$ to $2$, and multiply the result by $2$.
About $x=2$: Cylindrical shells seem like a good idea, but we can also do it by taking slices perpendicular to the $y$-axis.
Again we get a washer. The outer radius is the distance from $x=-\sqrt{4-4y^2}$ to $x=2$, and the inner radius is the distance from $\sqrt{4-4y^2}$ to $2$. When we set up the calculation, we get a lot of cancellation, and the result is $$\int_{y=-1}^1 16\pi\sqrt{1-y^2}\,dy.$$ Again, we can take advantage of symmetry, integrate from $0$ to $1$ and double the result.
The answers happen to be the same. We can see this by making the substitution $x=2t$ in the first integral.