Applications of Integration - Examples

On this page we display several examples to illustrate the concepts and theorems discussed in the Applications of Integration theme.

Problem
Let \[f(x)=x\mathrm{e}^{\frac{1}{2}x}+4.\] The area in the \(xy\)-plane bounded by the \(x\)-axis, the graph of \(f(x)\) and the lines \(x=\) and \(x=2\) is rotated around the \(y\)-axis. Find the volume of the resulting solid of revolution.

Solution
To calculate the volume by slicing, we need to solve \(y=f(x)\) for \(x\) and express the result as something that is easy to integrate. This is not so easy in our case. On the other hand, we observe that \(xf(x)\) can be integrated, and so it is better to use the method of cylindrical shells. The area is then

\[A=2\pi\int_0^2 x^2\mathrm{e}^{\frac{1}{2}x}+4x\, dx. \] We use integration by parts, twice, to find

\[A=2\pi\left( [x^2 2\mathrm{e}^{\frac{1}{2}x}]_0^2-\int_0^2 2x2\mathrm{e}^{\frac{1}{2}x}\, dx+[2x^2]_0^2\right)=\pi\left( [x^2 2\mathrm{e}^{\frac{1}{2}x}-8x\mathrm{e}^{\frac{1}{2}x}+16\mathrm{e}^{\frac{1}{2}x}+ 2x^2]_0^2\right)=2\pi(8\mathrm{e}-8).\]

Problem
Suppose a bowl can be modelled by rotating the graph of the function \(f(x)=kx^2\) on the interval \([0,a]\) around the \(y\)-axis. Given that the bowl should contain a fixed volume \(V\), we want to choose \(a>0\) and \(k>0\) such that the surface area is minimal.

Solution
The volume in question is the one resulting from rotating area below \(y=ka^2\) and above the graph \(f(x)\) around the \(y\)-axis. This can be divided into thin, horizontal, circular slices with thickness \(dy\) and radius \(x\), where \(x\) is such that \(f(x)=y\), which means that \(x=\sqrt{y/k}\). Thus the volume is given by

\[V=\pi\int_0^{ka^2}\left(\sqrt{\frac{y}{k}}\right)^2\, dy=\frac{\pi}{k}\int_0^{ka^2} y\, dy =\frac{\pi}{2k} [y^2]_0^{ka^2}=\frac{\pi ka^4}{2}.\]

The surface area can be found by dividing the area into small horizontal strips with radius \(2\pi r\) and width \(ds\). The radius is \(x\), and \(ds= \sqrt{1+\left(f'(x)\right)^2}dx\), so the surface area is then given by

\[ A=2\pi\int_0^a x\sqrt{1+4k^2x^2}\,dx=\frac{\pi}{6k^2}[(1+4k^2x^2)^{3/2}]_0^a=\frac{\pi (1+4k^2a^2)^{3/2}}{6k^2}-\frac{\pi}{6k^2}. \]

Using the relation \(k=\frac{2V}{\pi a^4}\) and inserting this into the last equation, we find a expression for \(A\) in terms of \(a\). We observe that both the volume and surface area grows more rapidly as \(a\) increases than when \(k\) increases. Then the optimal value for \(a\) is found by differentiating \(A\) w.r.t. \(a\) and finding the smallest positive root. This is easily done using maple.

Problem
Find the volume of the solid given by rotating the region \(R\) above the \(x\)-axis and below the graph of the function \(f(x)=4-(x-4)^2\) around the \(y\)-axis.

Solution
According to Pappus's theorem, the volume is given by \(V=2\pi \bar{r} A\), where \(\bar{r}\) is the perpendicular distance from the centroid of the region \(R\), and \(A\) is the area of \(R\). By symmetry, the perpendicular distance is the \(x\) coordinate of the centroid. We therefore calculate

\[V=2\pi\bar{r}A=2\pi\int_2^6 -x^3+8x^2-12x\,dx=2\pi\left[-\frac{x^4}{4}+\frac{8}{3}x^3-6x^2\right]_2^6=\frac{256}{3}\pi.\]