The region in the plane given by $$\left |{\frac {x}{2} - \frac {1}{2 \sqrt {\frac {x^{2}}{3} + 1}}}\right | \leq y \leq \frac {x}{2} + \frac {1}{2 \sqrt {\frac {x^{2}}{3} + 1}}$$ and $0 \leq x \leq 3$ is revolved around the $x$-axis. Compute the volume of the resulting solid.

If the variable $x$ is used for slicing, then slices are perpendicular to the axis of rotation, which indicates the washer method should be used. The inequalities for $y$ give the outer and inner radii, and $$\left (\frac {x}{2} + \frac {1}{2 \sqrt {\frac {x^{2}}{3} + 1}}\right )^2 - \left ( \left |{\frac {x}{2} - \frac {1}{2 \sqrt {\frac {x^{2}}{3} + 1}}}\right |\right )^2 = \frac {\sqrt {3} x}{\sqrt {x^{2} + 3}}.$$ (Note that the absolute values go away when the radius is squared.) This leads to the conclusion $$V= \int _{0}^{3} \frac {\sqrt {3} \pi x}{\sqrt {x^{2} + 3}} dx.$$ Among the options below, the best choice for a potential substitution is

Using the substitution above, the differentials satisfy $du = \answer {2x}dx$. This gives the equality $$\int \frac {\sqrt {3} \pi x}{\sqrt {x^{2} + 3}}\, dx = \int \answer {\frac {\sqrt {3} \pi }{2 \sqrt {u}}} du = \answer {\sqrt {3} \pi \sqrt {u}}.$$ Reversing the substitution gives $$\begin {aligned} \int _{0}^{3} \frac {\sqrt {3} \pi x}{\sqrt {x^{2} + 3}}\, dx & = \left . \left [\answer {\sqrt {3} \pi \sqrt {x^{2} + 3}} \right ] \right |_{0}^{3}\\ & = \answer {3 \pi }. \end {aligned}$$ Note: We would not have to reverse the substitution if we also determined the new bounds. In this case, if $u = x^2 + 3$ and $x = 0$, then $u = \answer {3}$. Likewise if $x = 3$, then $u = \answer {12}$. Thus we could also have carried out the calculation by changing bounds: $$\int _{0}^{3} \frac {\sqrt {3} \pi x}{\sqrt {x^{2} + 3}}\, dx = \int _{\answer {3}}^{\answer {12}} \answer {\frac {\sqrt {3} \pi }{2 \sqrt {u}}} du.$$

Using the table of integrals below, compute the indefinite integral $$\int \frac {d \theta }{(1-\theta ) \sqrt {\theta ^2 - 2 \theta - 3}}.$$

• When dealing with quadratic expressions such as the ones appearing in this integrand, it is often necessary to complete the square before appealing to a table. In this case, $$\theta ^2 - 2 \theta - 3 = \left ( \theta - \answer {1} \right )^2 - \answer {4}.$$
• Using the most appropriate entry of the table and plugging in the correct value of $a$ gives $$\int \frac {dx}{x \sqrt {x^2 - \answer {4}}} = \frac {1}{\answer {2}} \mathrm {arcsec} \frac {x}{\answer {2}} + C.$$
• Based on the results of completing the square, we make a substitution $x = \theta - \answer {1}$ and conclude $$\int \frac {d \theta }{(1-\theta ) \sqrt {\theta ^2 - 2 \theta - 3}} = \int \frac {dx}{\answer {-x} \sqrt {x^2 - \answer {4}}} = \answer {- \frac {1}{2}} \mathrm {arcsec} \answer {\frac {\theta -1}{2}} + C.$$

Use a table of integrals to compute the antiderivative below. $$\int \frac {x^2 ~ dx}{\sqrt {16 + x^6}}$$

• The key is to make the substitution $u = x^{\answer {3}}$ so that the expression $16 + x^6$ can be understood as a quadratic function of $u$.
• Specifically $du = \answer {3x^2} dx$, so $$\int \frac {x^2 ~ dx}{\sqrt {16 + x^6}} = \frac {1}{3} \int \frac {du}{\sqrt {16+u^2}}.$$ Aside from the factor of $1/3$, the $u$ integral belongs to the table: $$\int \frac {du}{\sqrt {16+u^2}} = \answer {\ln | u + \sqrt {16 + u^2}| } + C$$
• We conclude that $$\int \frac {x^2 ~ dx}{\sqrt {16 + x^6}} = \frac {1}{3} \int \frac {du}{\sqrt {16+u^2}} = \frac {\ln ( x^3 + \sqrt {16 + x^6})}{3} + C.$$ Absolute values are not needed in the logarithm because $x^3 + \sqrt {16 + x^6}$ can never be negative.