# Find the volume of solid inside the paraboloid z = 9 - x^2 - y^2, outside the cylinder x^2 + y^2...

## Question:

Find the volume of solid inside the paraboloid {eq}z = 9 - x^2 - y^2 {/eq}, outside the cylinder {eq}x^2 + y^2 = 4 {/eq}, and above the {eq}xy {/eq}-plane.

## Integration

A mathematical tool used to summarise the infinite small data or function in one form analyse the system is known as integration. It is widely used in the engineering field to design and analyse the system.

Given Data

• The equation of paraboloid is: {eq}z = 9 - {x^2} - {y^2}\cdots\cdots\rm{(I)} {/eq}
• {eq}{x^2} + {y^2} = 4 {/eq}

The expression for radius of curvature in polar coordinate is

{eq}{R^2} = {x^2} + {y^2} {/eq}

Subsitute and solve the above expression

{eq}\begin{align*} {R^2} &= 4\\ R &= 2 \end{align*} {/eq}

The domain in polar coordinate is

{eq}\begin{align*} 0 \le R \le 2\\ 0 \le \theta \le 2\pi \end{align*} {/eq}

Subsitute and solve expression (I)

{eq}\begin{align*} z &= 9 - \left( {{x^2} + {y^2}} \right)\\ &= 9 - {R^2} \end{align*} {/eq}

The expression for volume of solid inside the paraboloid is

{eq}\begin{align*} {V_o} &= \int_0^{2\pi } {\int_0^2 {\left( {9 - {R^2}} \right)RdRd\theta } } \\ &= \int_0^{2\pi } {\int_0^2 {\left( {9R - {R^3}} \right)dRd\theta } } \\ &= \left[ \theta \right]_0^{2\pi }\int_0^2 {\left( {9R - {R^3}} \right)dR} \\ &= \left[ {2\pi - 0} \right]\left[ {\dfrac{{9{R^2}}}{2} - \dfrac{{{R^4}}}{4}} \right]_0^2\\ &= 2\pi \left[ {\dfrac{{9{{\left( 2 \right)}^2}}}{2} - \dfrac{{{{\left( 2 \right)}^4}}}{4} - 0} \right]\\ &= 2\pi \times 14\\ &= 28\pi \end{align*} {/eq}

Thus the volume of paraboloid is {eq}28\pi {/eq}