$y=x \sqrt{4-x^2}$.
The region bounded by the curve between $x=2$ and $x=a$, where $0<a<2$, is rotated through $2\pi$ radians about the $x$-axis to generate a solid revolution with volume $\frac{657\pi}{160}$.
The question is to find the value of $a$. This is my working so far;
$\pi\int_a^2 (x\sqrt{4-x^2})^2=\frac{657\pi}{160}$
$\int_a^2 (x\sqrt{4-x^2})^2=\frac{657}{160}$
$[\frac{4}{3}x^3-\frac{1}{5}x^5]_a^2=\frac{657}{160}$
$\frac{657}{160}-\frac{64}{15}+\frac{4}{3}a^3-\frac{1}{5}a^5=0$
which gives $\frac{4}{3}a^3-\frac{1}{5}a^5-\frac{77}{480}=0$
or $640a^3-96a^5-77=0$
My question is, what method would be the best method to solve for $a$ in an exam and could you explain how to do so. Just to clarify, I know the answer, just not how to get to it.
We have $-96a^5+640a^3−77=0 \rightarrow 96a^5-640a^3+77=0$.
Because the leading term is even, we immediately attempt to factor the quintic into something of the form
$$(2a + n_0) (48 a^4 + n_1 a^3 + n_2 a^2 + n_3 a^3 + n_4) = 0$$
A fortunate case would be when the constant term is achieved with $\pm 1 \cdot \pm 77$. We recall that $0 < a < 2$, so we choose $-1 \cdot -77$ so that our simple factor, i.e. $(2a - 1)$, will actually be plausible.
This gives $$(2a \color{blue}{- 1}) (48 a^4 + n_1 a^3 + n_2 a^2 + n_3 a^3 \color{blue}{- 77}) = 0$$
Filling in the coefficients (which are uniquely determined after choosing $(2a - 1)$ as our first factor) gives
$$(2 a-1) (48 a^4 + 24a^3 - 308a^2 - 154a - 77) = 0$$
so our guess worked out! Since the root given by our easy factor ($a = \frac 12$) meets our condition that $0 < a < 2$, we are finished.
This was a very lucky quintic, so if things weren't so easy, we would have had to use the Rational Roots Theorem for our root candidates. (Note that there is no easy general solution for the quintic equation.)