Question

Use a double integral in polar coordinates to find the volume of the solid bounded by the graphs of the equations. $$ z=x^{2}+y^{2}+3, z=0, x^{2}+y^{2}=1 $$

Short answer

The final result after performing the double integration will be the volume of the solid.

Step-by-step solution

Converting to Polar Coordinates:

The given equations are in Cartesian coordinates. To use a double integral in polar coordinates, the equations need to be converted into polar coordinates. The conversion formulas are: \(x = r cos\theta \equal to 1, y = r sin\theta \equal to 1, z = r^{2} + 3 and z = 0\). The equation of the cylinder \(x^{2} + y^{2} = 1\) becomes \(r^{2} = 1\).

Setting up the Limits of Integration:

As the cylinder's equation became \(r^{2} = 1\), so \(r\) will vary from 0 to 1. Also, the object is symmetric to y-axis, so the variable \(\theta\) will range from 0 to \(2\pi\). Thus, the limits of the double integral are 0 to 1 for \(r\) and 0 to \(2\pi\) for \(\theta\). The \(z\) boundaries 0 and \(r^{2} + 3\) represent the height of the volume. To set up volume of the solid, the function to be integrated will be \(r^{2} + 3 - 0 = r^{2} + 3\). The area of a polar coordinate is given by \(dA = rdrd\theta\), so the differential \(dA\) needs to be included.

Performing the Double Integration:

The double integral, representing the volume of the solid, would look like this:\[V = \int_0^{2\pi} \int_0^1 (r^{2} + 3)r drd\theta\]Next, perform the inner integral, which will result in: \[V = \int_0^{2\pi} [\frac{1}{4}+3] d\theta \]Lastly, perform the outer integral to get the final answer.