Question

Find the volume of the given solid by subtracting two time the solid enclosed by the parabolic cylinders \(y = 1 - {x^2},z = {x^2} - 1\)and the planes \(x + y + z = 2,2x + 2y - z + 10 = 0\).

Short answer

The volume of the given solid can be:

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

Step-by-step solution

Find the limits

Consider the cylinders \(y = 1 - {x^2},z = {x^2} - 1\)

Consider the planes\(x + y + z = 2,2x + 2y - z + 10 = 0\)

Two of the boundaries are normal to the plane\(z = 0\)so the region of integration should be in the plane\(z = 0\)and using z for the integrand.

In order to set up the integral it’s helpful to graph the region of integration.

It’s a type 1 region so the graph includes a vertical guide line in the middle.

The sketch of the region is shown below:

Notice that the vertical line begins at\(y = {x^2} - 1\)and ends at \(y = 1 - {x^2}\).

The line could slide as far left as\(x = - 1\)and as far right as \(x = 1\).

Postponing the question of the integration momentarily, the form and limits of integration can be set up as: \(v = \int\limits_{ - 1}^1 {\int\limits_{{x^2} - 1}^{1 - {x^2}} {({\mathop{\rm int}} egrand)dydx} } \)

The desired volume is bounded above and below by planes.

Solve each of the two bounding planes for z.

One id the plane\({z_1} = 2x + 2y + 10\)and second and other is\({z_2} = 2 - x - y\)

Note that on the region of integration.

Therefore the desired volume can be computed by independently computing the volume beneath each plane and then subtracting.

The formula is: \(v = \int\limits_{ - 1}^1 {\int\limits_{{x^2} - 1}^{1 - {x^2}} {{z_1}dydx} } - \int\limits_{ - 1}^1 {\int\limits_{{x^2} - 1}^{1 - {x^2}} {{z_2}dydx} } \)

Find the volume \({V_1}\) bounded by \({z_1}\)

\(\begin{aligned}{l}{V_1} = \int\limits_{ - 1}^1 {\int\limits_{{x^2} - 1}^{1 - {x^2}} {2x + 2y + 10dydx} } \\{V_1} = \int\limits_{ - 1}^1 {\left[ {2xy + 2\frac{{{y^2}}}{2} + 10y} \right]_{{x^2} - 1}^{1 - {x^2}}} dx\\{V_1} = \int\limits_{ - 1}^1 {\left[ {\left( {2x + 10} \right)y + {y^2}} \right]_{{x^2} - 1}^{1 - {x^2}}} dx\\{V_1} = \int\limits_{ - 1}^1 {\left[ {\left( {2x + 10} \right)\left( {1 - {x^2}} \right) + {{\left( {1 - {x^2}} \right)}^2} - \left( {2x + 10} \right)\left( {{x^2} - 1} \right) - {{\left( {{x^2} - 1} \right)}^2}} \right]} dx\\{V_1} = \int\limits_{ - 1}^1 {\left[ {\left( {2x - 2{x^3} + 10 - 10{x^2} + 1 - 2{x^2} + {x^4} - \left( {2{x^3} - 2x + 10{x^2} - 10 + {x^4} - 2{x^2} + 1} \right)} \right)} \right]} dx\\{V_1} = \int\limits_{ - 1}^1 {\left[ {\left( { - 4{x^3} - 20{x^2} + 4x + 20} \right)} \right]} dx\\{V_1} = 2\int\limits_0^1 {\left[ {\left( { - 20{x^2} + 20} \right)} \right]} dx\\{V_1} = 40\left[ { - \frac{1}{3}{x^3} + x} \right]_0^1\\{V_1} = \frac{{80}}{3}\end{aligned}\)

Find the volume \({V_2}\) bounded by \({z_2}\)

\(\begin{aligned}{l}{V_2} = \int\limits_{ - 1}^1 {\int\limits_{{x^2} - 1}^{1 - {x^2}} {2 - x - ydydx} } \\{V_2} = \int\limits_{ - 1}^1 {\left[ {2y - xy - \frac{{{y^2}}}{2}} \right]_{{x^2} - 1}^{1 - {x^2}}dx} \\{V_2} = \int\limits_{ - 1}^1 {\left[ {\left( {2\left( {1 - {x^2}} \right) - x\left( {1 - {x^2}} \right) - \frac{{{{\left( {1 - {x^2}} \right)}^2}}}{2}} \right) - \left( {2\left( {{x^2} - 1} \right) - x\left( {{x^2} - 1} \right) - \frac{{{{\left( {{x^2} - 1} \right)}^2}}}{2}} \right)} \right]dx} \\{V_2} = \int\limits_{ - 1}^1 {\left[ {\left( {2 - 2{x^2} - x + {x^3} - \frac{1}{2} + {x^2} - \frac{1}{2}{x^4} - \left( { - 2 + 2{x^2} + x - {x^3} - \frac{1}{2} - {x^2} + \frac{1}{2}{x^4}} \right)} \right)} \right]dx} \\{V_2} = \int\limits_{ - 1}^1 {\left[ {\left( {2{x^3} - 4{x^2} - 2x + 4} \right)} \right]dx} \\{V_2} = 2\int\limits_0^1 {\left[ {\left( { - 4{x^2} + 4} \right)} \right]dx} \\{V_2} = 8\left[ { - \frac{1}{3}{x^3} + x} \right]_0^1\\{V_2} = \frac{{16}}{3}\end{aligned}\)

Find the volume

\(\begin{aligned}{l}V = \frac{{80}}{3} - \frac{{16}}{3}\\V = \frac{{80 - 16}}{3}\\V = \frac{{64}}{3}\end{aligned}\)

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