Question

Find the volume of the solid generated by revolving the region bounded by the curves = 4 and = 2 about the -axis.

Answer 1

Volume of solid of Revolution

    $V = \pi \int_0 ^b y^{2} dx$

      $V = \pi \int_0^4 (4x^2)^2 dx\\ \\ = 16\pi \int_0^4 x^4 dx$

Now we can integrate and evaluate:

    $V = \frac{16\pi x^5}{5} |_0^4$

         $= \frac{16\pi(4)^5}{5}$

    V = $\frac{16384\pi}{5} \approx 10294.3$

   

Answer 2

Answer:

The volume of solid Revolution,

$V=\frac{16384 \pi}{5} \approx 10294.3$

Step-by-step explanation:

Given :

The region bounded by the curves = 4 and = 2 about the -axis.

To find:

The volume of the solid generated by revolving the region bounded by the curves = 4 and = 2 about the -axis.

Step 1

The volume of solid Revolution,

$\mathrm{V}=\int_{0}^{1} \pi \mathrm{x}^{2} \mathrm{dy}$

$V=\pi \int_{0}^{b} x^{2} d x$

$=16\pi \int_{0}^{4} x^{4} d x$

$=\pi \int_{0}^{4} (4x^{2})^{2} d x$

Step 2

Integrating the above equation

$V=\left.\frac{16 \pi x^{5}}{5}\right|_{0} ^{4}$

$=\left.\frac{16 \pi (4)^{5}}{5}$

$V=\frac{16384 \pi}{5}$

$\approx 10294.3$

Therefore, the volume of solid Revolution

$V=\frac{16384 \pi}{5} \approx 10294.3$