Paul's Online Notes
Home / Calculus II / Applications of Integrals / Surface Area
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 2-2 : Surface Area

3. Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating $$y = {x^3} + 4$$ , $$1 \le x \le 5$$ about the given axis. You can use either $$ds$$.

1. the $$x$$-axis
2. the $$y$$-axis
Show All Solutions Hide All Solutions

a $$x$$-axis Show All Steps Hide All Steps
Start Solution

We are told that we can use either $$ds$$ here and the function seems to be set up to use the following $$ds$$.

$ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$

Note that we could use the other $$ds$$ if we wanted to. However, that would require us to solve the equation for $$x$$ in terms of $$y$$. That would, in turn, would give us fractional exponents that would make the derivatives and hence the integral potentially messier.

Therefore, we’ll go with our first choice of $$ds$$.

Show Step 2

Now we’ll need the derivative of the function.

$\frac{{dy}}{{dx}} = 3{x^2}$

Plugging this into the formula for our choice of $$ds$$ gives,

$ds = \sqrt {1 + {{\left[ {3{x^2}} \right]}^2}} \,dx = \sqrt {1 + 9{x^4}} \,dx$ Show Step 3

Finally, all we need to do is set up the integral. Also note that we have a $$dx$$ in the formula for $$ds$$ and so we know that we need $$x$$ limits of integration which we’ve been given in the problem statement.

$SA = \int_{{}}^{{}}{{2\pi y\,ds}} = \int_{1}^{5}{{2\pi y\,\sqrt {1 + 9{x^4}} \,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{1}^{5}{{2\pi \left( {{x^3} + 4} \right)\,\sqrt {1 + 9{x^4}} \,dx}}}}$

Be careful with the formula! Remember that the variable in the integral is always opposite the axis of rotation. In this case we rotated about the $$x$$-axis and so we needed a $$y$$ in the integral.

Finally, with the $$ds$$ we choose to use for this part we had a $$dx$$ in the final integral and that means that all the variables in the integral need to be $$x$$’s. This means that the $$y$$ from the formula needs to be converted into $$x$$’s as well. Luckily this is easy enough to do since we were given the formula for $$y$$ in terms of $$x$$ in the problem statement.

b $$y$$-axis Show All Steps Hide All Steps
Start Solution

We are told that we can use either $$ds$$ here and the function seems to be set up to use the following $$ds$$ for the same reasons we choose it in the first part.

$ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$ Show Step 2

Now, as with the first part of this problem we’ll need the derivative of the function and the $$ds$$. Here is that work again for reference purposes.

$\frac{{dy}}{{dx}} = 3{x^2}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}ds = \sqrt {1 + {{\left[ {3{x^2}} \right]}^2}} \,dx = \sqrt {1 + 9{x^4}} \,dx$ Show Step 3

Finally, all we need to do is set up the integral. Also note that we have a $$dx$$ in the formula for $$ds$$ and so we know that we need $$x$$ limits of integration which we’ve been given in the problem statement.

$SA = \int_{{}}^{{}}{{2\pi x\,ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{1}^{5}{{2\pi x\,\sqrt {1 + 9{x^4}} \,dx}}}}$

Be careful with the formula! Remember that the variable in the integral is always opposite the axis of rotation. In this case we rotated about the $$y$$-axis and so we needed an $$x$$ in the integral.

In this part, unlike the first part, we do not do any substitution for the $$x$$ in front of the root. Our choice of $$ds$$ for this part put a $$dx$$ into the integral and this means we need $$x$$’s the integral. Since the variable in front of the root was an $$x$$ we don’t need to do any substitution for the variable.