Question

Find the volume of the given solid by subtracting two time the solid enclosed by the parabolic cylinders \(y = {x^2}\)and the planes\(z = 3y,z = 2 + y\).

Short answer

The volume of the given solid can be:

\(V = \frac{{16}}{{15}}\).

Step-by-step solution

Find the limits

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

Consider the planes\(z = 3y,z = 2 + y\)

We set\(z = 3y\)and\(z = 2 + y\)equal to each other and solve for y

\(\begin{aligned}{l}3y = 2 + y\\2y = 2\\y = 1\end{aligned}\)

We have the solution\(y = 1\)and since\({x^2} \ge 0,y \ge 0\)which means that\({x^2} \le y \le 1\)

We set\(y = 1\)and\(y = {x^2}\)equal to each other and solve for x

We have the solution\(x = \pm 1\)which means that\( - 1 \le x \le 1\)

We can subtract two equations of the plane in any manner we choose, but if the resulting integration is negative, we take the absolute value of the integral to get the volume.

In this instance we construct\(f(x,y)\)as follows:

\(\begin{aligned}{l}f(x,y) = 2 + y - (3y)\\f(x,y) = 2 - 2y\end{aligned}\)

The limit of the integration can be given as:

\( - 1 \le x \le 1,{x^2} \le y \le 1\)

The formula is: \(V = \int\limits_{ - 1}^1 {\int\limits_{{x^2}}^1 {\left( {2 - 2y} \right)dydx} } \)

Find the volume \(V\)

\(\begin{aligned}{l}V = \int\limits_{ - 1}^1 {\int\limits_{{x^2}}^1 {\left( {2 - 2y} \right)dydx} } \\V = \int\limits_{ - 1}^1 {\left( {2y - {y^2}} \right)_{{x^2}}^1dx} \\V = \int\limits_{ - 1}^1 {\left( {2 - 1} \right) - \left( {2{x^2} - {x^4}} \right)dx} \\V = \int\limits_{ - 1}^1 {1 - 2{x^2} + {x^4}dx} \\V = \left[ {x - \frac{{2{x^3}}}{3} + \frac{{{x^5}}}{5}} \right]_{ - 1}^1\\V = \left[ {\left( {1 - \frac{2}{3} + \frac{1}{5}} \right) - \left( { - 1 + \frac{2}{3} - \frac{1}{5}} \right)} \right]\\V = \frac{{16}}{{15}}\end{aligned}\)

Hence, The volume of the given solid can be evaluated as: \(V = \frac{{16}}{{15}}\).