Question

Find the volume of the given solid. Bounded by the cylinders \({x^2} + {y^2} = 1\) and the planes \(y = x,x = 0,z = 0\)in the first octant.

Short answer

The volume of the given solid can be:

\(V = \frac{1}{3}\).

Step-by-step solution

Find the limits

Consider the cylinders \({x^2} + {y^2} = 1\)

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

We can write \({x^2} + {y^2} = 1 \Rightarrow z = \pm \sqrt {1 - {y^2}} \)

The region is in first octant so the limit of x is: \(0 \le x \le \sqrt {1 - {y^2}} \)

the limit of y is: \(0 \le y \le 1\)

The sketch of the region is shown below:

From the graph, The region of integration is shown below: \(D = \{ (x,y)|0 \le y \le 1,0 \le x \le \sqrt {1 - {y^2}} \} \)

The volume can be given by:

Find the integral

Take\(u = 1 - {y^2}\)

\(\begin{aligned}{l}du = - 2ydy\\ - \frac{1}{2}du = ydy\end{aligned}\)

For\(y = 0\)

\(u = 1 - {y^2} \Rightarrow u = 1\)

For\(y = 1\)

\(u = 1 - {y^2} \Rightarrow u = 0\)

Substitute \(u = 1 - {y^2}\)and \( - \frac{1}{2}du = ydy\) and simplify

\(\begin{aligned}{l}V = \int\limits_0^1 {y\sqrt {1 - {y^2}} (dy)} \\V = \int\limits_1^0 { - \sqrt u ( - \frac{1}{2}du)} \\V = \frac{1}{2}\int\limits_1^0 {\sqrt u (du)} \\V = \frac{1}{2}\left[ {\frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^1\\V = \frac{1}{3}\left[ {{u^{\frac{3}{2}}}} \right]_0^1\\V = \frac{1}{3}\left[ {{1^{\frac{3}{2}}}} \right]\\V = \frac{1}{3}\end{aligned}\)

Hence, The volume of the given solid can be evaluated as: \(V = \frac{1}{3}\).