Question

Find the volume of the given solid. Enclosed by the paraboloid \(z = {x^2} + 3{y^2}\)and the planes \(x = 0,y = 1,y = x,z = 0\)

Short answer

The volume of the given solid can be:

\(V = \frac{5}{6}\).

Step-by-step solution

Find the limits

Consider the paraboloid \(z = {x^2} + 3{y^2}\)

The planes are\(x = 0,y = 1,y = x,z = 0\)

The region of integration is shown below:

From the figure observe that x varies from\(0\)to\(y\)and y varies from\(0\)to\(2\).

The volume can be given by: \(V = \int\limits_0^1 {\int\limits_0^y {\left( {{x^2} + 3{y^2}} \right)dx} } dy\)

Find the integral

\(\begin{aligned}{l}V = \int\limits_0^1 {\int\limits_0^y {\left( {{x^2} + 3{y^2}} \right)dx} } dy\\V = \int\limits_0^1 {\left( {\frac{{{x^3}}}{3} + 3{y^2}x} \right)_0^y} dy\\V = \int\limits_0^1 {\left( {\frac{{{y^3}}}{3} + 3{y^3}} \right)} dy\\V = \left( {\frac{{{y^4}}}{{12}} + \frac{{3{y^4}}}{4}} \right)_0^1\\V = \left( {\frac{{{1^4}}}{{12}} + \frac{{3{{(1)}^4}}}{4} - 0} \right)\\V = \frac{1}{{12}} + \frac{3}{4}\\V = \frac{5}{6}\end{aligned}\)

Hence, The volume of the given solid can be evaluated as:\(V = \frac{5}{6}\).