1.6 Volumes

The horizontal cross-sections were discussed in Example 1.6.1 of the CLP-2 text.

The horizontal cross-sections are circles, but the vertical cross-sections are not.

What are the dimensions of the cross-sections?

The columns have the same volume.

There are two different kinds of washers.

Draw sketches. The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which is in the optional section 1.6 of the CLP-2 text. The method of washers also works, but requires you to have more patience and also to have a good idea what the specified region looks like. Look at your sketch very careful when identifying the ends of your horizontal strips.

(a) $\pi\displaystyle\int_{0}^{3} xe^{2x^2}\ \dee{x}$

(b) $\displaystyle\int _0^1 \pi\big[\big(3+\sqrt{y}\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} +\displaystyle\int _ 1^4 \pi\big[\big(5-y\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y}$

Draw sketchs.

(a) $\displaystyle\int_{-1}^{1}\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big]\,\dee{x}$ (b) $\displaystyle\int _{-1}^0 \pi\big[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\big]\,\dee{y}$

Draw a sketch.

$\pi\displaystyle\int_{-2}^{2}\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\ \dee{x}$

If you take horizontal slices (parallel to one face), they will all be equilateral triangles.

Be careful not to confuse the height of a triangle with the height of the tetrahedron.

$\dfrac{\sqrt{2}}{12}\ell^3$

Sketch the region.

$\displaystyle\frac{\pi}{4}\Big(e^{2a^2}-1\Big)$

Sketch the region first.

$\pi\left[\dfrac{38}{3}-\dfrac{514}{3^4}\right] = \pi\dfrac{512}{81}$

You can save yourself quite a bit of work by interpreting the integral as the area of a known geometric figure.

(a) $8\pi\int_{-1}^1\sqrt{1-x^2}\,\dee{x}$ (b) $4\pi^2$

See Example 1.6.3 in the

CLP-2 text.

(a) The region $R$ is the region between the blue and red curves, with $3\le x\le 5$, in the figures below.

(b) $\frac{4}{3}\pi\approx 4.19$

See Example 1.6.5 in the

CLP-2 text.

(a) The region $R$ is sketched below.

(b) $\pi\Big[4\log 2 - \frac{3}{2}\Big] \approx 3.998$

Sketch the region. To find where the curves intersect, look at where $\cos(\frac x2)$ and $x^2 - \pi^2$ both have roots.

$\pi^2 + 8\pi^3 + \frac{8\pi^6}{5}$

See Example 1.6.6 in the

CLP-2 text.

$\dfrac{8}{3}$

See Example 1.6.6 in the

CLP-2 text. Imagine cross-sections with shadow parallel to the $y$-axis, sticking straight out of the $xy$-plane.

$\dfrac{256\times 8}{15}=136.5\dot3$

See Example 1.6.1 in the

CLP-2 text.

$\dfrac{28}{3}\pi h$

(a) Don't be put off by phrases like “rotating an ellipse about its minor axis." This is the same kind of volume you've been calculating all section.
(b) Hopefully, you sketched the ellipse in part (a). What was its smallest radius? Its largest? These correspond to the polar and equitorial radii, respectively.
(c) Combine your answers from (a) and (b).
(d) Remember that the absolute error is the absolute difference of your two results–that is, you subtract them and take the absolute value. The relative error is the absolute error divided by the actual value (which we're taking, for our purposes, to be your answer from (c)). When you take the relative error, lots of terms will cancel, so it's easiest to not use a calculator till the end.

(a) $\dfrac{4\pi}{3b^2a}$ cubic units (b) $a = \dfrac{1}{6356.752}$ and $b=\dfrac{1}{6378.137}$

(c) Approximately $1.08321\times 10^{12} ~\mathrm{km}^3$, or $1.08321\times 10^{21}~ \mathrm{m}^3$

(d) Absolute error is about $3.64\times 10^{9}~ \mathrm{km}^3$, and relative error is about $0.00336$, or $0.336\%$.

To find the points of intersection, set $4-(x-1)^2=x+1$.

(a) $\dfrac{9}{2}$ (b) $\pi\displaystyle\int_{-1}^2 \big[{\big(4-x\big)}^2-{\big(1+(x-1)^2\big)}^2\big]\,\dee{x}$

You can somewhat simplify your calculations in part (a) (but not part (b)) by using the fact that $\cR$ is symmetric about the line $y=x$.

When you're solving an equation for $x$, be careful about your signs: $x-1$ is negative.

(a) $\dfrac{\pi}{2}-1$ (b) $\dfrac{\pi^2}{2}-\pi\approx 1.793$

The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what $\cR$ looks like. So it is crucial to first sketch $\cR$. Then be very careful in identifying the left end of your horizontal strips.

(a) $V_1=\dfrac{4}{3}\pi c^2$ (b) $V_2 =\dfrac{\pi\,c}{3}\big[4\sqrt{2}-2 \big]$ (c) $c=0\text{ or }c=\sqrt{2}-\frac{1}{2}$

Note that the curves cross. The area of this region was found in Problem 14 of Section 1.5. It would be useful to review that problem.

$\displaystyle\int_{\pi/2}^\pi \pi\big[(5 + \pi \sin x)^2-(5 + 2\pi - 2x)^2\big]\ \dee{x} +\displaystyle\int^{3\pi/2}_\pi \pi\big[(5 + 2\pi - 2x)^2-(5 + \pi \sin x)^2\big]\ \dee{x}$

You can use ideas from this section to answer the question. If you take a very thin slice of the column, the density is almost constant, so you can find the mass. Then you can add up all your little slices. It's the same idea as volume, only applied to mass.

Do be careful about units: in the problem statement, some are given in metres, others in kilometres.

If you're having a hard time with the antiderivative, try writing the exponential function with base $e$. Remember $2 = e^{\log 2}$.

(a) $\dfrac{6000c\pi}{\log 2}\left(1-\dfrac{1}{2^{10}}\right)$, which is close to $\dfrac{6000c\pi}{\log 2}$.

(b) 6km: that is, there is roughly the same mass of air in the lowest 6 km of the column as there is in the remaining 54 km.