# How do you find the volume bounded by y=x^2, x=y^2 revolved about the x=-1?

Jul 2, 2016

$= \frac{29}{30} \pi$

#### Explanation:

Consider the small element width dx as shown, being revolved around the line x = -1

The cross sectional area (csa) of the washer of width dx is the csa of the outer circle minus the csa of the inner

ie $\mathrm{dA} = \pi {\left(1 + x + \mathrm{dx}\right)}^{2} - \pi {\left(1 + x\right)}^{2}$

simplify the algebra with $u = 1 + x$ so we have

$\mathrm{dA} = \pi \left({\left(u + \mathrm{dx}\right)}^{2} - {u}^{2}\right) = \pi \left(2 u \mathrm{dx} + {\mathrm{dx}}^{2}\right)$
$= \pi \left(2 \left(1 + x\right) \mathrm{dx} + {\mathrm{dx}}^{2}\right)$

we can already see that $\frac{\mathrm{dA}}{\mathrm{dx}} {|}_{\mathrm{dx} \to 0} = \pi \left(2 \left(1 + x\right) \textcolor{red}{+ \mathrm{dx}}\right)$

so we can ignore $m a t h c a l \left(O\right) \left({\mathrm{dx}}^{2}\right)$ in the original expression

thus

$\mathrm{dA} = 2 \pi \left(1 + x\right) \mathrm{dx}$

the volume of that small element is

$\mathrm{dV} = \left({y}_{2} - {y}_{1}\right) \mathrm{dA}$

$= \left({y}_{2} - {y}_{1}\right) \cdot 2 \pi \left(1 + x\right) \mathrm{dx}$

$= 2 \pi \left(1 + x\right) \left(\sqrt{x} - {x}^{2}\right) \mathrm{dx}$

$\setminus \implies V = 2 \pi {\int}_{0}^{1} \mathrm{dx} q \quad \left(1 + x\right) \left(\sqrt{x} - {x}^{2}\right)$

$= 2 \pi {\int}_{0}^{1} \mathrm{dx} q \quad \left(\sqrt{x} - {x}^{2} + {x}^{\frac{3}{2}} - {x}^{3}\right)$

$= 2 \pi {\left(\frac{2}{3} {x}^{\frac{3}{2}} - {x}^{3} / 3 + \frac{2}{5} {x}^{\frac{5}{2}} - {x}^{4} / 4\right)}_{0}^{1}$

$= 2 \pi \left(\frac{2}{3} - \frac{1}{3} + \frac{2}{5} - \frac{1}{4}\right)$

$= \frac{29}{30} \pi$