Calculus II - Surface Area

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Section 8.2 : Surface Area

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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 $ds$ .

the $x$ -axis
the $y$ -axis

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$x$ -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.

$y$ -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

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

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.