5.2: Volumes with Cross Sections

The concept of volume is already familiar to us. We now elaborate on its definition: Suppose you make a straight cut through a solid three-dimensional object. The region through which you slice is a two-dimensional plane, which we call a cross section. Making a straight cut through a sphere's center, for example, yields a circular cross section. The volume of any solid with uniform cross-sectional area \(A\) and height \(h\) is given by the formula \[V = A h \pd\] For example, a cylinder of radius \(r\) and height \(h\) has uniform cross-sectional area \(A = \pi r^2,\) so its volume is \(V = \pi r^2 h.\) Likewise, a rectangular prism of length \(\ell,\) width \(w,\) and height \(h\) contains a uniform cross-sectional area of \(A = \ell w,\) meaning its volume is \(V = \ell w h.\) (See Figure 1.)

Nonuniform Cross Sections Let's calculate the volume of a solid \(S\) whose cross sections are not uniform. From \(x = a\) to \(x = b\) let the continuous function \(A(x)\) be the area of the cross section to \(S\) in a plane \(P_x\) that is perpendicular to the \(x\)-axis at any point \(x.\) Note that the solid's cross-sectional area \(A(x)\) changes with \(x.\) For example, at the endpoint \(a\) the cross section to \(S\) perpendicular to the \(x\)-axis has area \(A(a).\) Likewise, at \(b\) the cross section to \(S\) perpendicular to the \(x\)-axis has area \(A(b).\) (See Figure 2.) Imagine cutting \(S\) into \(n\) slabs of equal width \(\Delta x,\) similar to slicing a loaf of bread. Over the interval \([a, b],\) these slabs form endpoints \(a = x_0,\) \(x_1, \dots, x_{n - 1}, x_n = b.\) Choosing any sample point \(x_i^*\) in the general subinterval \([x_{i - 1}, x_i],\) the \(i\)th slab of \(S\)—which has cross-sectional area \(A(x_i^*)\) and height \(\Delta x\)—has a volume of \[\Delta V = A(x_i^*) \Delta x \pd\] Summing the volumes of all \(n\) slabs, we approximate the volume of \(S\) to be \[ \ba V \approx \sum_{i = 1}^n A(x_i^*) \Delta x \pd \ea \] As \(n\) increases, the many thinner slabs better approximate the volume \(V.\) So \(V\) is the limiting value of the sum as \(n \to \infty\)—that is, \[V = \lim_{n \to \infty} \sum_{i = 1}^n A(x_i^*) \Delta x \pd\] The right side is a Riemann sum for the function \(A(x),\) so we find \begin{equation} V = \int_a^b A(x) \di x \pd \label{eq:volume-A(x)} \end{equation} In words, if we are given a solid's cross-sectional area \(A\) at every \(x\) in \([a, b],\) then we calculate the solid's volume by integrating \(A(x)\) from \(x = a\) to \(x = b.\)

Let's think of the \(xy\)-plane as the ground, the base of the solid \(S\) shown in Figure 3. Because cross sections perpendicular to the \(x\)-axis are squares, the cross-sectional area of the solid at any \(x\) is \(A(x) = \par{4 - x^2}^2.\) Figure 4 shows an approximating slice of \(S,\) whose volume is \[\Delta V = A(x) \Delta x = \par{4 - x^2}^2 \Delta x \pd\] By \(\eqref{eq:volume-A(x)},\) the solid's volume \(V\) is given by integrating \(A(x)\) from \(x = 0\) to \(x = 2,\) as follows: \[ \ba V &= \int_0^2 A(x) \di x = \int_0^2 \par{4 - x^2}^2 \di x \nl &= \int_0^2 \par{16 - 8x^2 + x^4} \di x \nl &= \par{16x - \tfrac{8}{3} x^3 + \tfrac{1}{5} x^5} \intEval_0^2 \nl &= \boxed{\tfrac{256}{15}} \approx 17.067 \pd \ea \]

At each \(x,\) the cross section of the solid \(S\) (shown in Figure 5) is a rectangle of dimensions \(y = \sqrt x\) and \(2y = 2 \sqrt x.\) Thus, at each \(x\) the cross-sectional area of the solid is \[A(x) = \par{\sqrt x \,} \par{2 \sqrt x \,} = 2 x \pd\] An approximating rectangular prism shown by Figure 6 has volume \[\Delta V = A(x) \Delta x = 2x \Delta x \pd\] By \(\eqref{eq:volume-A(x)},\) the volume of \(S\) is given by integrating \(A(x)\) from \(x = 0\) to \(x = 4 \col\) \[ \ba V &= \int_0^4 A(x) \di x = \int_0^4 2x \di x \nl &= x^2 \intEval_0^4 = \boxed{16} \ea \]

At each \(x,\) the cross section of the solid \(S\) (shown in Figure 7) is a right isosceles triangle whose hypotenuse has length \(y = \sqrt{9 - x^2}.\) Accordingly, the triangle's legs each have length \(y/\sqrt 2.\) So at each \(x,\) the cross-sectional area of the solid is \[A(x) = \tfrac{1}{2} \par{\frac{y}{\sqrt 2}}^2 = \frac{y^2}{4} = \frac{9 - x^2}{4} \pd\] (See Figure 8.) By \(\eqref{eq:volume-A(x)},\) the volume of \(S\) is given by integrating \(A(x)\) from \(x = -3\) to \(x = 3 \col\) \[ V = \int_{-3}^3 A(x) \di x = \int_{-3}^3 \frac{9 - x^2}{4} \di x \pd \] But using symmetry (since the integrand is even), we simplify the calculation as follows: \[ \ba V &= 2 \cdot \int_0^3 \frac{9 - x^2}{4} \di x = \tfrac{1}{2} \int_0^3 (9 - x^2) \di x \nl &= \tfrac{1}{2} \par{9x - \tfrac{1}{3} x^3} \intEval_0^3 = \boxed 9 \ea \]

The ellipse \(4x^2 + y^2 = 9\) has vertices at \((0, -3)\) and \((0, 3).\) At each \(y,\) the cross section to the semi-ellipsoid \(S\) (shown in Figure 9) perpendicular to the \(y\)-axis is a semicircle. Hence, each semicircle's diameter is \(2x\)—where \(x\) is a function of \(y\)—so its radius is \(x\) and its area is \(A(y) = \tfrac{1}{2} \pi x^2.\) (See Figure 10.) Solving for the positive solution of \(x\) in the equation of the ellipse gives \[x = \tfrac{1}{2} \sqrt{9 - y^2} \pd\] So we have \[A(y) = \tfrac{1}{2} \pi \par{\tfrac{1}{2} \sqrt{9 - y^2}}^2 = \frac{\pi}{8} (9 - y^2) \pd\] By \(\eqref{eq:volume-A(x)},\) the semi-ellipsoid's volume is given by integrating \(A(y)\) from \(y = -3\) to \(y = 3\)—that is, \[ V = \int_{-3}^3 A(y) \di y = \int_{-3}^3 \frac{\pi}{8} \par{9 - y^2} \di y \pd \] But through symmetry (because the integrand is even), it is easier to compute the volume as follows: \[ \ba V &= 2 \cdot \int_0^3 \frac{\pi}{8} \par{9 - y^2} \di y \nl &= \frac{\pi}{4} \int_0^3 \par{9 - y^2} \di y \nl &= \frac{\pi}{4} \par{9y - \tfrac{1}{3} y^3} \intEval_0^3 \nl &= \boxed{\frac{9 \pi}{2}} \approx 14.137 \pd \ea \]

Let's rewrite \(y = 2 - x\) as \(x = 2 - y.\) The enclosed region \(R\) is bounded in the first and fourth quadrants, and the two graphs intersect at \((1, 1)\) and \((4, -2).\) Accordingly, a plane that intersects the solid \(S\) (shown in Figure 11) perpendicular to the \(y\)-axis at any \(y\) yields a semicircle whose diameter has length \[(2 - y) - y^2 = 2 - y - y^2 \pd\] Thus, the semicircle's radius \(r\) is half this value—meaning the solid's cross-sectional area at any \(y\) is \[A(y) = \tfrac{1}{2} \pi r^2 = \tfrac{1}{2} \pi \par{\frac{2 - y - y^2}{2}}^2 = \tfrac{1}{8} \pi \par{2 - y - y^2}^2 \pd\] (See Figure 12.) So following \(\eqref{eq:volume-A(x)},\) we find the solid's volume to be \[ \ba V &= \int_{-2}^1 A(y) \di y = \int_{-2}^1 \tfrac{1}{8} \pi \par{2 - y - y^2}^2 \di y \nl &= \frac{\pi}{8} \int_{-2}^1 \par{y^4 + 2y^3 - 3y^2 - 4y + 4} \di y \nl &= \frac{\pi}{8} \par{\tfrac{1}{5} y^5 + \tfrac{1}{2} y^4 - y^3 - 2y^2 + 4y} \intEval_{-2}^1 \nl &= \boxed{\frac{81\pi}{80}} \approx 3.181 \pd \ea \]