# What is the volume of the solid produced by revolving f(x)=sqrt(81-x^2) around the x-axis?

##### 1 Answer

Volume $\textcolor{b l u e}{V = 972 \pi \text{ }}$cubic units

#### Explanation:

Solution 1.

The given curve is located at the first and second quadrants as shown in the graph

You will notice that the graph shows that it is a half circle with radius $r = 9 \text{ }$units. If we revolve this about the x-axis the solid form is a sphere.

Formula for volume of the sphere

$V = \frac{4}{3} \pi {r}^{3}$

$V = \frac{4}{3} \pi {9}^{3}$

$V = \frac{4}{3} \pi 729$

$\textcolor{b l u e}{V = 972 \pi \text{ }}$cubic units

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Solution 2.

To solve for the volume by the Calculus , we make use of the Disk Method

$\mathrm{dV} = \pi {r}^{2} \mathrm{dh}$

$\mathrm{dV} = \pi {y}^{2} \mathrm{dx}$

$V = {\int}_{- 9}^{9} \pi \cdot {y}^{2} \mathrm{dx} = \pi \cdot {\int}_{- 9}^{9} {\sqrt{\left(81 - {x}^{2}\right)}}^{2} \mathrm{dx}$

$V = \pi \cdot {\int}_{- 9}^{9} \left(81 - {x}^{2}\right) \mathrm{dx}$

$V = \pi \cdot {\left[81 x - {x}^{3} / 3\right]}_{- 9}^{9}$

$V = \pi \cdot \left[81 \cdot 9 - {9}^{3} / 3 - \left(81 \left(- 9\right) - {\left(- 9\right)}^{3} / 3\right)\right]$

$V = \pi \cdot \left[729 - \frac{729}{3} - \left(- 729 + \frac{729}{3}\right)\right]$

$V = \pi \cdot \left[729 - 243 + 729 - 243\right]$

$\textcolor{b l u e}{V = 972 \pi \text{ }}$cubic units.

God bless....I hope the explanation is useful.