Question

Find the volume of the following solid regions. The solid bounded by the paraboloid \(z=x^{2}+y^{2}\) and the plane \(z=9\)

Short answer

Answer: The volume of the solid region is \(\dfrac{405\pi}{2}\) cubic units.

Step-by-step solution

Convert to cylindrical coordinates

First, let's convert the given functions into cylindrical coordinates. The cylindrical coordinate system uses coordinates \((r,\theta,z)\), which are related to the Cartesian coordinates \((x,y,z)\) by $$x = r\cos{\theta}$$ $$y = r\sin{\theta}$$ $$z = z$$ The given paraboloid equation is \(z=x^2+y^2\). Replacing \(x\) and \(y\) with their cylindrical coordinates counterparts, we get: $$z= (r\cos{\theta})^2 + (r\sin{\theta})^2$$ $$z = r^2(\cos^2{\theta} + \sin^2{\theta})$$ Since \(\cos^2{\theta} + \sin^2{\theta} = 1\), we now have $$z = r^2$$

Set up the triple integral

To find the volume of the solid region, we need to set up a triple integral in cylindrical coordinates. The volume element in cylindrical coordinates is \(dV = r\,dr\,d\theta\,dz\). The limits of integration for \(r\), \(\theta\), and \(z\) will be determined by the paraboloid equation and the plane equation. The plane equation is \(z = 9\). From the paraboloid equation in cylindrical coordinates, we have \(z = r^2\). The solid is bounded above by the plane and below by the paraboloid, so the limits for \(z\) will be from \(r^2\) to \(9\). The limits for \(r\) will be from \(0\) to \(\sqrt{9}\), and for \(\theta\), it will be from \(0\) to \(2\pi\). The triple integral to be evaluated is then: $$V=\int_{0}^{2\pi}\int_{0}^{\sqrt{9}}\int_{r^2}^9 r\, dz\, dr\, d\theta$$

Evaluate the triple integral

Now, let's evaluate the triple integral. We start with the innermost integral: $$\int_{r^2}^9 r\, dz = r[z]_{r^2}^9 = 9r - r^3$$ Now, let's evaluate the middle integral: $$\int_{0}^{\sqrt{9}} (9r - r^3)\, dr = \big[ \dfrac{9}{2}r^2 - \dfrac{1}{4}r^4 \big]_{0}^{\sqrt{9}}= \dfrac{9}{2}(9)(9) - \dfrac{1}{4}(9)^2 = \dfrac{243}{2} - \dfrac{81}{4}=\dfrac{405}{4}$$ Finally, evaluate the outermost integral: $$\int_{0}^{2\pi}\big(\dfrac{405}{4}\big) d\theta = \dfrac{405}{4}[\theta]_0^{2\pi}=\dfrac{405\pi}{2}$$

Write the final answer

The volume of the solid region bounded by the paraboloid \(z=x^2+y^2\) and the plane \(z=9\) is \(\dfrac{405\pi}{2}\) cubic units.