Volumes

If R is a region in the x-y plane and $z = f(x, y)$ is a function, the volume of the solid lying above R and below the graph of the function is given by

$$\dint_R f(x, y)\,dx\,dy.$$

The picture gives a heuristic justification for this:

$$\hbox{\epsfysize=2.5in \epsffile{volumes-1.eps}}$$

The region R is partitioned into boxes, each $dx$ by $dy$ . Above a box we construct a rectangular parallelepiped (i.e. a "tall box") up to the surface. The height of the box is $z = f(x, y)$ , the height of the surface. The volume of a "tall box" is $f(x, y)\,dx\,dy$ . The double integral "adds up" the volumes of the "tall boxes" over R to get the total volume.

A careful justification would use Riemann sums for the double integral.

This is really a signed volume: If the function is zero or negative on R, the integral may not represent the physical volume.

Example. Find the volume of the solid lying below the graph of $z = 24 x^2 y^2$ and above the following region in the x-y-plane:

$$\hbox{\epsfysize=1.5in \epsffile{volumes-2.eps}}$$

The region is

$$\left\{\matrix{ 1 \le x \le 2 \cr -1 \le y \le 1 \cr}\right\}$$

The volume is

$$\int_{-1}^1 \int_1^2 24 x^2 y^2\,dx\,dy = \int_{-1}^1 \left[8 x^3 y^2\right]_1^2\,dy = \int_{-1}^1 56 y^2\,dy = \left[\dfrac{56}{3} y^3\right]_{-1}^1 = \dfrac{112}{3} = 37.33333 \ldots.\quad\halmos$$

Example. Find the volume of the solid lying below the graph of $z = 18 x y$ and above the following region in the x-y-plane:

$$\hbox{\epsfysize=1.5in \epsffile{volumes-3.eps}}$$

First, I'll find the equation of the line, which has x-intercept 3 and y-intercept 2. Suppose the line is $a x + b y = c$ . The x-intercept is $(3, 0)$ , so plugging this in, I get

$$3 a = c, \quad\hbox{so}\quad a = \dfrac{c}{3}.$$

The y-intercept is $(0, 2)$ , so plugging this in I get

$$2 b = c, \quad\hbox{so}\quad b = \dfrac{c}{2}.$$

Thus,

$$\eqalign{ \dfrac{c}{3} x + \dfrac{c}{2} y & = c \cr \noalign{\vskip2pt} 2 x + 3 y & = 6 \cr \noalign{\vskip2pt} y & = \dfrac{1}{3} (6 - 2 x) \cr}$$

The triangular region is

$$\left\{\matrix{ 0 \le x \le 3 \cr \noalign{\vskip2pt} 0 \le y \le \dfrac{1}{3} (6 - 2 x) \cr}\right\}$$

The volume is

$$\int_0^3 \int_0^{(6 - 2 x)/3} 18 x y\,dy\,dx = \int_0^3 \left[9 x y^2\right]_0^{(6 - 2 x)/3}\,dx = \int_0^3 x (6 - 2 x)^2\,dx = \int_0^3 (36 x - 24 x^2 + 4 x^3)\,dx =$$

$$\left[18 x^2 - 8 x^3 + x^4\right]_0^2 = 27.\quad\halmos$$

Example. Find the volume of the solid lying below the graph of $z = 3 x y^2$ and above the following region in the x-y-plane:

$$R = \left\{\matrix{ 0 \le x \le 1 \cr 0 \le y \le 1 - x^2 \cr}\right\}$$

$$\hbox{\epsfysize=1.5in \epsffile{volumes-4.eps}}$$

$$\int_0^1 \int_0^{1 - x^2} 3 x y^2\,dy\,dx = \int_0^1 \left[x y^3\right]_0^{1 - x^2}\,dx = \int_0^1 x (1 - x^2)^3\,dx = \int_1^0 x u^3 \cdot \dfrac{du}{-2 x} =$$

$$\left[u = 1 - x^2, \quad du = -2 x\,dx, \quad dx = \dfrac{du}{-2 x}; \quad x = 0, u = 1; x = 1, u = 0\right]$$

$$\dfrac{1}{2} \int_0^1 u^3\,du = \dfrac{1}{2} \left[\dfrac{1}{4} u^4\right]_0^1 = \dfrac{1}{8}.\quad\halmos$$

If $f(x, y) \ge g(x, y)$ for $(x, y)$ in a region R, the volume bounded above by the graph of f and bounded below by the graph of g, and lying inside the cylinder determined by R, is given by

$$\dint_R [f(x, y) - g(x, y)]\,dx\,dy.$$

Example. Find the volume of the solid bounded above by $z = 4 y + 1$ and bounded below by $z = -1 - 2 x$ , and lying inside the triangular cylinder

$$\left\{\matrix{ 0 \le y \le 1 \cr 0 \le x \le 1 - y \cr}\right\}.$$

$$\hbox{\epsfysize=1.5in \epsffile{volumes-5.eps}}$$

The volume is

$$\int_0^1 \int_0^{1 - y} \left((4 y + 1) - (-1 - 2 x)\right)\,dx\,dy = \int_0^1 \int_0^{1 - y} \left(2 x + 4 y + 2\right)\,dx\,dy = \int_0^1 \left[x^2 + 4 x y + 2 x\right]_0^{1 - y}\,dy =$$

$$\int_0^1 \left((1 - y)^2 + 4 y (1 - y) + 2 (1 - y)\right)\,dy = \int_0^1 (-3 y^2 + 3)\,dy =$$

$$\left[-y^3 + 3 y\right]_0^1 = 2.\quad\halmos$$

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