§6.2 Using definite integrals to find volume

Introduction

The Volume of a Solid of Revolution

Revolving about the

A cone of radius $3$ and height $5$ lies along the $x$-axis with its base at $x=0$. The line forming its top edge is

The region bounded by $y = 4-x^2$ and the $x$-axis is revolved about the $x$-axis. Using disks of radius $4-x^2$ on $[-2,2]$, the exact volume is $\frac{512\pi}{15}$. Report the decimal, rounded to three places.

The region between $y = 4-x^2$ and $y = x+2$, which intersects where $x^2+x-2 = 0$, is revolved about the $x$-axis. The correct washer setup is

The region bounded by the $x$-axis, $y = \sqrt{x}$, and $x = 4$ is revolved about the $x$-axis. The volume equals

The region bounded by the $y$-axis, $y = \sqrt{x}$, and $y = 2$ is revolved about the $x$-axis. The volume equals

The region between $y = \sqrt{x}$ and $y = x^3$, which intersect at $(0,0)$ and $(1,1)$, is revolved about the $x$-axis. Using washers, the exact volume is $\frac{5\pi}{14}$. Report the decimal, rounded to three places.

The region between $y = 2x^2+1$ and $y = x^2+4$ on $[-\sqrt{3},\sqrt{3}]$ is revolved about the $x$-axis. The volume integrand is

Using the washer setup for $y = 2x^2+1$ and $y = x^2+4$ about the $x$-axis, the exact volume is $\frac{136\sqrt{3}}{5}\pi$. Report the decimal, rounded to three places.

The region bounded by the $y$-axis, $y = \sqrt{x}$, and $y = 2$ is revolved about the $y$-axis, giving $V = \int_0^2\pi y^4\,dy$. Report the volume, rounded to three decimals.

The region between $y = \sqrt{x}$ and $y = x^4$ is revolved about the $y$-axis. Writing the curves as $x = y^2$ and $x = y^{1/4}$, the washer setup is

The region bounded by the $x$-axis, $y = \sqrt{x}$, and $x = 4$ is revolved about the $y$-axis. The correct setup and volume are

The first quadrant region between $y = 2x$ and $y = x^3$ is revolved about the $x$-axis, giving $V = \int_0^{\sqrt{2}}\pi(4x^2-x^6)\,dx$. The exact value is $\frac{32\sqrt{2}\pi}{21}$. Report the decimal, rounded to three places.

The same first quadrant region between $y = 2x$ and $y = x^3$ is revolved about the $y$-axis. Writing $x = y^{1/3}$ and $x = \frac{1}{2}y$, the washer setup is

The region bounded by $x = (y-1)^2$ and $y = x-1$, rewritten as $x = y+1$, is revolved about the $y$-axis. The washer setup is

When the region between $y = 2x$ and $y = x^3$ is revolved about the $y$-axis, the washer method requires rewriting both curves as functions of $y$ and integrating with respect to $y$.