Question

Find the volume of the given solid. Bounded by the cylinders \({y^2} + {z^2} = 4\) and the planes \(x = 2y,x = 0,z = 0\)in the first octant.

Short answer

The volume of the given solid can be:

\(V = \frac{{16}}{3}\).

Step-by-step solution

Find the limits

Consider the cylinders \({y^2} + {z^2} = 4\)

Planes are: \(x = 2y,x = 0,z = 0\)

We can write \({y^2} + {z^2} = 4 \Rightarrow z = \sqrt {4 - {y^2}} \)

The region enclosed by\(x = 0\)and \(x = 2y\)where \(0 \le y \le 2\)

The sketch of the region bounded by the curves \(z = \sqrt {4 - {y^2}} ,x = 0\)and \(x = 2y\) is shown below:

From the graph, The region of integration is shown below: \(D = \{ (x,y)|0 \le x \le 2y,0 \le y \le 2\} \)

The volume can be given by:

Find the integral

Take\(u = \sqrt {4 - {y^2}} \)

\(\begin{aligned}{l}du = \frac{{ - 2ydy}}{{2\sqrt {4 - {y^2}} }}\\du = \frac{{ - ydy}}{{\sqrt {4 - {y^2}} }}\\du = \frac{{ - ydy}}{u}\\ - udu = ydy\end{aligned}\)

For\(y = 0\)

\(u = \sqrt {4 - {y^2}} \Rightarrow u = 2\)

For\(y = 2\)

\(u = \sqrt {4 - {y^2}} \Rightarrow u = 0\)

substitute \(u = \sqrt {4 - {y^2}} \)and \( - udu = ydy\) and simplify

\(\begin{aligned}{l}V = 2\int\limits_0^2 {\sqrt {4 - {y^2}} (ydy)} \\V = 2\int\limits_0^2 {u( - udu)} \\V = 2\int\limits_0^2 {{u^2}du} \\V = 2\left[ {\frac{{{u^3}}}{3}} \right]_0^2\\V = 2\left[ {\frac{{{2^3}}}{3}} \right]\\V = \frac{{16}}{3}\end{aligned}\)

Hence, The volume of the given solid can be evaluated as:\(V = \frac{{16}}{3}\).