Surface Area Of A Function

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Section 2-ii : Surface Area

In this section nosotros are going to expect once once again at solids of revolution. We first looked at them back in Calculus I when nosotros found the volume of the solid of revolution. In this section nosotros want to find the surface area of this region.

So, for the purposes of the derivation of the formula, allow's await at rotating the continuous function $y = f\left( x \correct)$ in the interval $\left[ {a,b} \correct]$ nearly the $x$-axis. We'll as well need to presume that the derivative is continuous on $\left[ {a,b} \right]$. Beneath is a sketch of a office and the solid of revolution we get by rotating the function about the $x$-axis.

$This is the graph of some unknown office on the domain a<10<b. It is completely in the 1st quadrant and looks like a parabola that opens upwards. The left end of the graph is lower than right end of the graph.$ This is the graph of the solid we get from rotating the graph from above about the x-axis. It looks like an odd vase laying on its side. It has a wide

Nosotros can derive a formula for the surface surface area much every bit we derived the formula for arc length. Nosotros'll start past dividing the interval into $northward$ equal subintervals of width $\Delta x$. On each subinterval nosotros volition judge the function with a straight line that agrees with the function at the endpoints of each interval. Here is a sketch of that for our representative office using $northward = 4$.

$This is a graph of the function above on the domain a<x<b. The domain is split into 4 equal subintervals and the corresponding points on the graph are labeled $P_{0}$, $P_{1}$, $P_{2}$, $P_{3}$ and $P_{4}$. A line then connects each of these points approximating the graph of the curve in each subinterval.$

Now, rotate the approximations about the $ten$-axis and we get the following solid.

The solid of revolution in the second graph above is approximated by a series of

The approximation on each interval gives a distinct portion of the solid and to make this clear each portion is colored differently. Each of these portions are chosen frustums and we know how to observe the surface area of frustums.

The surface area of a frustum is given by,

\[A = 2\pi rl\]

where,

\[\brainstorm{align*}r = \frac{1}{2}\left( {{r_1} + {r_2}} \right)\hspace{0.25in}\hspace{0.25in}{r_1} = & {\mbox{radius of right end}}\\ \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{r_2} = & {\mbox{radius of left terminate}}\end{align*}\]

and $fifty$ is the length of the slant of the frustum.

For the frustum on the interval $\left[ {{x_{i - 1}},{x_i}} \correct]$ nosotros have,

\[\begin{marshal*}{r_1} = & f\left( {{x_i}} \correct)\\ {r_2} = & f\left( {{x_{i - 1}}} \correct)\\ l = & \left| {{P_{i - one}}\,\,{P_i}} \right|\,\,\,\,\,\,\left( {{\mbox{length of the line segment connecting }}{P_i}{\mbox{ and }}{P_{i - 1}}} \correct)\end{align*}\]

and we know from the previous section that,

\[\left| {{P_{i - one}}\,\,{P_i}} \right| = \sqrt {1 + {{\left[ {f'\left( {x_i^*} \correct)} \correct]}^two}} \,\,\,\Delta x\,\,\,\,\,{\mbox{where }}x_i^*{\mbox{ is some point in }}\left[ {{x_{i - i}},{x_i}} \right]\]

Earlier writing down the formula for the surface expanse we are going to assume that $\Delta x$ is "pocket-size" and since $f\left( 10 \right)$ is continuous we tin and then assume that,

\[f\left( {{x_i}} \right) \approx f\left( {x_i^*} \right)\hspace{0.25in}\hspace{0.25in}{\mbox{and}}\hspace{0.25in}\hspace{0.25in}f\left( {{x_{i - one}}} \right) \approx f\left( {x_i^*} \right)\]

So, the surface area of the frustum on the interval $\left[ {{x_{i - 1}},{x_i}} \right]$ is approximately,

\[\brainstorm{align*}{A_{\,i}} & = 2\pi \left( {\frac{{f\left( {{x_i}} \correct) + f\left( {{x_{i - 1}}} \right)}}{2}} \correct)\left| {{P_{i - 1}}\,\,{P_i}} \correct|\,\,\\ & \approx 2\pi f\left( {x_i^*} \right)\sqrt {ane + {{\left[ {f'\left( {x_i^*} \right)} \right]}^two}} \,\,\,\Delta x\end{align*}\]

The surface area of the whole solid is and so approximately,

\[Southward \approx \sum\limits_{i = one}^n {2\pi f\left( {x_i^*} \right)\sqrt {1 + {{\left[ {f'\left( {x_i^*} \right)} \correct]}^2}} \,\,\,\Delta x} \]

and we tin can become the exact surface area by taking the limit as $north$ goes to infinity.

\[\begin{align*}South & = \mathop {\lim }\limits_{north \to \infty } \sum\limits_{i = 1}^due north {2\pi f\left( {x_i^*} \right)\sqrt {1 + {{\left[ {f'\left( {x_i^*} \right)} \correct]}^2}} \,\,\,\Delta x} \\ & = \int_{{\,a}}^{{\,b}}{{two\pi f\left( x \right)\sqrt {ane + {{\left[ {f'\left( ten \right)} \right]}^2}} \,dx}}\finish{align*}\]

If nosotros wanted to nosotros could as well derive a like formula for rotating $x = h\left( y \right)$ on $\left[ {c,d} \correct]$ near the $y$-axis. This would give the post-obit formula.

\[S = \int_{{\,c}}^{{\,d}}{{2\pi \,h\left( y \right)\sqrt {1 + {{\left[ {h'\left( y \right)} \correct]}^2}} \,dy}}\]

These are not the "standard" formulas however. Notice that the roots in both of these formulas are zero more than than the two $ds$'s we used in the previous section. Also, we will replace $f\left( x \right)$ with $y$ and $h\left( y \right)$ with $x$. Doing this gives the following two formulas for the surface area.

Surface Surface area Formulas

\[\begin{array}{ll}\begin{align*}South = \int{{ii\pi y\,ds}}\hspace{0.25in}\hspace{0.25in}{\mbox{rotation most }}x - {\mbox{axis}}\\ S = \int{{2\pi x\,ds}}\hspace{0.25in}\hspace{0.25in}{\mbox{rotation almost }}y - {\mbox{axis}}\end{align*}\end{array}\]

where,

\[\begin{assortment}{ll}\brainstorm{marshal*}ds = \sqrt {one + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx\,\hspace{0.25in}{\mbox{if }}y = f\left( x \right),\,\,a \le ten \le b\\ ds = \sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \correct)}^2}} \,dy\,\hspace{0.25in}{\mbox{if }}10 = h\left( y \correct),\,\,c \le y \le d\cease{marshal*}\stop{array}\]

In that location are a couple of things to note well-nigh these formulas. First, notice that the variable in the integral itself is ever the opposite variable from the one nosotros're rotating about. Second, nosotros are allowed to use either $ds$ in either formula. This means that there are, in some mode, iv formulas here. We will choose the $ds$ based upon which is the most convenient for a given part and trouble.

Now permit'south work a couple of examples.

Example i Make up one's mind the surface surface area of the solid obtained by rotating $y = \sqrt {9 - {x^2}} $, $ - 2 \le ten \le 2$ nigh the $x$-centrality.

Show Solution

The formula that we'll be using here is,

\[S = \int{{ii\pi y\,ds}}\]

since we are rotating about the $x$-centrality and we'll employ the get-go $ds$ in this example because our part is in the correct form for that $ds$ and we won't proceeds anything by solving it for $x$.

Permit'due south get-go go the derivative and the root taken care of.

\[\frac{{dy}}{{dx}} = \frac{one}{2}{\left( {9 - {x^ii}} \right)^{ - \frac{1}{2}}}\left( { - 2x} \right) = - \frac{x}{{{{\left( {9 - {x^2}} \right)}^{\frac{1}{2}}}}}\] \[\sqrt {one + {{\left( {\frac{{dy}}{{dx}}} \correct)}^2}} = \sqrt {1 + \frac{{{ten^ii}}}{{ix - {ten^2}}}} = \sqrt {\frac{9}{{9 - {x^2}}}} = \frac{3}{{\sqrt {nine - {x^2}} }}\]

Hither's the integral for the surface surface area,

\[S = \int_{{\, - two}}^{{\,2}}{{2\pi y\frac{3}{{\sqrt {9 - {x^2}} }}\,dx}}\]

There is a problem even so. The $dx$ ways that we shouldn't have any $y$'southward in the integral. So, before evaluating the integral we'll need to substitute in for $y$ also.

The surface expanse is then,

\[\begin{align*}Southward & = \int_{{\, - 2}}^{{\,2}}{{2\pi \sqrt {9 - {10^two}} \frac{3}{{\sqrt {nine - {ten^2}} }}\,dx}}\\ & = \int_{{\, - 2}}^{{\,2}}{{6\pi \,dx}}\\ & = 24\pi \end{align*}\]

Previously we made the annotate that nosotros could use either $ds$ in the surface area formulas. Let's work an example in which using either $ds$ won't create integrals that are too difficult to evaluate and then we can check both $ds$'south.

Example two Determine the area of the solid obtained by rotating $y = \sqrt[iii]{x}$, $1 \le y \le 2$ about the $y$-centrality. Utilise both $ds$'due south to compute the expanse.

Show Solution

Note that we've been given the role set up up for the starting time $ds$ and limits that work for the second $ds$.

Solution ane
This solution will use the start $ds$ listed to a higher place. Nosotros'll offset with the derivative and root.

\[\frac{{dy}}{{dx}} = \frac{1}{3}{x^{ - \frac{2}{three}}}\] \[\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} = \sqrt {ane + \frac{i}{{9{x^{\frac{4}{3}}}}}} = \sqrt {\frac{{nine{x^{\frac{4}{three}}} + 1}}{{9{x^{\frac{4}{three}}}}}} = \frac{{\sqrt {9{10^{\frac{iv}{3}}} + i} }}{{3{x^{\frac{2}{3}}}}}\]

We'll as well need to get new limits. That isn't too bad however. All we need to practice is plug in the given $y$'due south into our equation and solve to get that the range of $x$'s is $ane \le 10 \le eight$. The integral for the expanse is then,

\[\begin{marshal*}S & = \int_{{\,one}}^{{\,eight}}{{2\pi x\frac{{\sqrt {nine{x^{\frac{iv}{3}}} + 1} }}{{3{x^{\frac{two}{3}}}}}\,dx}}\\ & = \frac{{ii\pi }}{3}\int_{{\,1}}^{{\,8}}{{{x^{\frac{1}{3}}}\sqrt {9{10^{\frac{4}{3}}} + 1} \,dx}}\stop{marshal*}\]

Notation that this time nosotros didn't demand to substitute in for the $x$ as we did in the previous example. In this case we picked upward a $dx$ from the $ds$ and and then we don't demand to practice a exchange for the $x$. In fact, if nosotros had substituted for $x$ we would have put $y$'s into the integral which would have caused bug.

Using the substitution

\[u = 9{ten^{\frac{4}{3}}} + 1\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}du = 12{x^{\frac{one}{three}}}\,dx\]

the integral becomes,

\[\begin{marshal*}S & = \frac{\pi }{{xviii}}\int_{{\,10}}^{{\,145}}{{\sqrt u \,du}}\\ & = \left. {\frac{\pi }{{27}}{u^{\frac{3}{2}}}} \correct|_{10}^{145}\\ & = \frac{\pi }{{27}}\left( {{{145}^{\frac{iii}{ii}}} - {{10}^{\frac{3}{2}}}} \right) = 199.48\stop{align*}\]

Solution 2
This time we'll use the second $ds$. So, we'll first need to solve the equation for $x$. We'll as well get alee and get the derivative and root while nosotros're at information technology.

\[x = {y^three}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\frac{{dx}}{{dy}} = 3{y^2}\] \[\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} = \sqrt {i + 9{y^4}} \]

The surface area is then,

\[Southward = \int_{{\,i}}^{{\,2}}{{ii\pi x\sqrt {1 + ix{y^four}} \,dy}}\]

We used the original $y$ limits this time because we picked up a $dy$ from the $ds$. Also note that the presence of the $dy$ ways that this time, dissimilar the start solution, we'll need to substitute in for the $x$. Doing that gives,

\[\begin{marshal*}S & = \int_{{\,one}}^{{\,ii}}{{2\pi {y^3}\sqrt {i + nine{y^4}} \,dy}}\hspace{0.25in}\hspace{0.25in}u = 1 + ix{y^4}\\ & = \frac{\pi }{{xviii}}\int_{{\,10}}^{{\,145}}{{\sqrt u \,du}}\\ & = \frac{\pi }{{27}}\left( {{{145}^{\frac{3}{two}}} - {{10}^{\frac{3}{two}}}} \right) = 199.48\cease{align*}\]

Note that after the substitution the integral was identical to the first solution and so the piece of work was skipped.

As this example has shown we tin can use either $ds$ to get the area. It is important to point out equally well that with one $ds$ we had to do a exchange for the $x$ and with the other we didn't. This will e'er piece of work out that way.

Note as well that in the case of the terminal example it was just as easy to use either $ds$. That ofttimes won't exist the case. In many examples merely i of the $ds$ volition be user-friendly to work with so we'll always need to determine which $ds$ is liable to be the easiest to piece of work with before starting the problem.