# Solids of revolution | Exercise 1

A solid of revolution is generated by rotating around the axis $y$, the region bounded by the curve $y=\sqrt[3]{x}$, the axis $x$ and the line $x = c$ where $c > 0$. Calculate the value of $c$ so that the solid has a volume of $12\pi u^{3}$.  The $u^{3}$ means cubic units.

To find the volume of the solid of revolution, we will use the method of cylindrical shells.

Let’s graph the function of $y=\sqrt[3]{x}$ and put the line $x=c$

We are going to write the equation with which we will find the value of $c$, what we will do is write an integral and we will match it to the volume value of $12\pi u^{3}$ that the exercise tells us, the limits of the integral must be from zero to the value of $c$:

$$V = \int_{0}^{c}2\pi x\sqrt[3]{x} = 12 \pi$$

By properties of the integrals we will extract the constants of the integral and reduce the $x$’s:

$$2\pi \int_{0}^{c} x^{\frac{4}{3}} = 12 \pi$$

We cancel terms and perform the integral of $x^{\frac{4}{3}}$ which, by chain rule, equals $\frac{3}{7}x^{\frac{7}{3}}$:

$$\require{cancel} \cancel{2}\cancel{\pi} \int_{0}^{2} x^{\frac{4}{3}} = \cancelto{6}{12} \cancel{\pi}$$

$$\left. \cfrac{3x^{\frac{7}{3}}}{7}\right]_{0}^{c} = 6$$

We evaluate, multiply by $7$ and divide by $3$ all equality:

$$\cfrac{3c^{\frac{7}{3}}}{7} = 6$$

$$c^{\frac{7}{3}} = 14$$

Next we will raise all the equality to the cube to eliminate the denominator $3$ of the power of $\frac{7}{3}$:

$$c^{7} = 14^{3} = 2744$$

Now we will apply the root to the $7$ to all the equality to obtain the value of $c$:

$$c = \sqrt[7]{2744} \approx 3.098$$

So our value of $c$ for the volume to be $12\pi u^{3}$, must have a value of $\sqrt[7]{2744}$.

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