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### Section 8.2 : 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
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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.