Solutions to Problem Set 14

Math 211-03

10-5-2017

[Volumes by cross-sections]

1. The cross-sectional area of a solid in planes perpendicular to the x-axis is

$$A(x) = x \sin 2 x \quad\hbox{for}\quad 0 \le x \le \dfrac{\pi}{2}.$$

Find the volume of the solid.

$$V = \int_0^{\pi/2} x \sin 2 x\,dx = \left[-\dfrac{1}{2} x \cos 2 x + \dfrac{1}{4} \sin 2 x\right]_0^{\pi/2} = \dfrac{\pi}{4} = 0.78539 \ldots.$$

Here's the work for the integral:

$$\matrix{& \der {} x & & \displaystyle \int\,dx \cr & & & \cr + & x & & \sin 2 x \cr & & \searrow & \cr - & 1 & & -\dfrac{1}{2} \cos 2 x \cr & & \searrow \cr + & 0 & & -\dfrac{1}{4} \sin 2 x \cr}$$

$$\int x \sin 2 x\,dx = -\dfrac{1}{2} x \cos 2 x + \dfrac{1}{4} \sin 2 x + c.\quad\halmos$$


2. The base of a solid is the region in the x-y plane bounded by $y = x$ and $y = x^2 -
   6$ . The cross-sections in planes perpendicular to the x-axis area squares having one edge in the x-y plane. Find the volume of the solid.

Find the intersection points:

$$\eqalign{ x^2 - 6 & = x \cr x^2 - x - 6 & = 0 \cr (x - 3)(x + 2) & = 0 \cr x = 3 & \quad\hbox{or}\quad x = -2 \cr}$$

$$\hbox{\epsfysize=1.5in \epsffile{volumes-by-cross-sections2.eps}}$$

A typical square cross-section has sides of length $x - (x^2 - 6) = 6 + x - x^2$ . The volume is

$$V = \int_{-2}^3 (6 + x - x^2)^2\,dx = \int_{-2}^3 \left(36 + 12 x - 11 x^2 - 2 x^3 + x^4\right)\,dx = \left[36 x + 6 x^2 - \dfrac{11}{3} x^3 - \dfrac{1}{2} x^4 + \dfrac{1}{5} x^5\right]_{-2}^3 =$$

$$\dfrac{625}{6} \approx 104.16667.\quad\halmos$$


3. The base of a solid is the semicircular region bounded by $y = \sqrt{4 - x^2}$ and the x-axis. The cross-sections in planes perpendicular to the x-axis are equilateral triangles having one edge in the x-y plane. Find the volume of the solid.

$$\hbox{\epsfysize=1.5in \epsffile{volumes-by-cross-sections3.eps}}$$

A typical triangular cross-section has sides of length $\sqrt{4 - x^2}$ . Since the area of an equilateral triangle with sides of length s is $\dfrac{s^2
   \sqrt{3}}{4}$ , the volume is

$$V = \int_{-2}^2 \dfrac{(\sqrt{4 - x^2})^2 \sqrt{3}}{4}\,dx = \dfrac{\sqrt{3}}{4} \int_{-2}^2 (4 - x^2)\,dx = \dfrac{\sqrt{3}}{4} \left[4 x - \dfrac{1}{3} x^3\right]_{-2}^2 = \dfrac{8 \sqrt{3}}{3} = 4.61880 \ldots.\quad\halmos$$


4. The base of a solid is the region in the x-y plane bounded by $y = 3 + 2 x - x^2$ and the x-axis. The cross-sections in planes perpendicular to the x-axis are isosceles right triangles with their hypotenuses lying the in x-y plane. Find the volume of the solid.

Find the x-intercepts:

$$\eqalign{ 3 + 2 x - x^2 & = 0 \cr x^2 - 2 x - 3 & = 0 \cr (x - 3)(x + 1) & = 0 \cr x = 3 & \quad\hbox{or}\quad x = -1 \cr}$$

$$\hbox{\epsfysize=1.5in \epsffile{volumes-by-cross-sections4.eps}}$$

A typical triangular cross-section has a hypotenuse of length $3 + 2 x - x^2$ . If the hypotenuse of an isosceles right triangle is h, the legs have length $\dfrac{h}{\sqrt{2}}$ . Therefore, the legs of the cross-section have length $\dfrac{3 + 2 x - x^2}{\sqrt{2}}$ . Therefore, the area of a cross-section is

$$\dfrac{1}{2} (\hbox{base}) (\hbox{height}) = \dfrac{1}{2} \left(\dfrac{3 + 2 x - x^2}{\sqrt{2}}\right) \left(\dfrac{3 + 2 x - x^2}{\sqrt{2}}\right) = \dfrac{1}{4} (3 + 2 x - x^2)^2.$$

The volume is

$$V = \int_{-1}^3 \dfrac{1}{4} (3 + 2 x - x^2)^2\,dx = \dfrac{1}{4} \int_{-1}^3 \left(9 + 12 x - 2 x^2 - 4 x^3 + x^4\right)\,dx =$$

$$\dfrac{1}{4} \left[9 x + 6 x^2 - \dfrac{2}{3}x^3 - x^4 + \dfrac{1}{5}x^5\right]_{-1}^3 = \dfrac{128}{15} \approx 8.53333.\quad\halmos$$


5. A solid hemisphere of radius 2 is cut by a plane 1 unit above and parallel to the base.

$$\hbox{\epsfysize=1.5in \epsffile{volumes-by-cross-sections5a.eps}}$$

The top piece is removed. What is the volume of the piece that remains?

If I slice the solid parallel to the base, I get circular cross-sections. Here is a side view:

$$\hbox{\epsfysize=1.5in \epsffile{volumes-by-cross-sections5b.eps}}$$

A typical circular cross-section is shown on edge in the picture. Its radius is

$$r = \sqrt{2^2 - y^2} = \sqrt{4 - y^2}.$$

Therefore, its area is

$$\pi r^2 = \pi(4 - y^2).$$

Thus, the volume is

$$\int_0^1 \pi(4 - y^2)\,dy = \pi\left[4y - \dfrac{1}{3}y^3\right]_0^1 = \dfrac{11\pi}{3} \approx 11.51917.\quad\halmos$$


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