Problem 71 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 $$z = r^2$$

Step 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: $$V=\int_{0}^{2\pi}\int_{0}^{\sqrt{9}}\int_{r^2}^9 r\, dz\, dr\, d\theta$$

Step 3: 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: $$\int_{0}^{2\pi}\big(\dfrac{405}{4}\big) d\theta = \dfrac{405}{4}[\theta]_0^{2\pi}=\dfrac{405\pi}{2}$$

Step 4: 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.