Find the volume of a right circular cone of height 14 whose base is a circle of radius 6.
You're going to need the cross-sectional area here. How could you find that? The cross-sections should be circles.
Use similar triangles to figure out what the radius of each circle is as a function of height.
$\displaystyle r = 6 - \frac{3h}{7}$. If this is $r$, what is the area of each cross section?
$\displaystyle V = \int_0^{14} \pi \left(6 - \frac{3h}{7}\right)^2\ dh$
$\displaystyle 168\pi$
Find the volume of the solid whose base is the region bounded between $y = x^2$ and $y=5$ and whose cross-sections perpendicular to the $y$-axis are squares.
Sketch out a picture. Which way are the cross-sections going, and how which variable should we integrate in?
What is the length of the base of each cross section in terms of $y$? Given this, what is the area as a function of $y$?
What should the bounds be on the integral that you need to compute?
The integral you will need to compute is $ \displaystyle \int_0^5 4x\ dx $
$\displaystyle 50$
Find the volume of the solid whose base is the circle $x^2 + y^2 = 1$, and whose cross-sections perpendicular to the $x$-axis are equilateral triangles.
What shape are the cross-sections? You're going to know the length of the base of this shape; try to find a formula for the area in terms of this length.
If you have an equilateral triangle of base L, the height will be $\displaystyle \frac{\sqrt{3}}{2}L$. (Prove it)
Since we have cross-sections perpendicular to the $x$ axis, this should be an integral in $x$. How long is the base of the triangle at position $x$?
The base is length $2\sqrt{1 - x^2}$, which gives an area of the triangle as $$ \frac{1}{2} L \left(\frac{\sqrt{3}}{2}L \right) = \sqrt{3}(1 - x^2)$$.
$\displaystyle \int_{-1}^1 \sqrt{3}(1-x^2)\ dx = \frac{4\sqrt{3}}{3}$
Find the total mass of a $2$ m rod whose linear density function is $\rho(x) = 10(x+5)^{-1}$ for $ 0 \leq x \leq 2$
This should be computed by an integral. What does it look like?
The mass can be computed by $ \displaystyle \int_0^2 \frac{10}{x+5}\ dx$.
$\displaystyle 10 \ln\left(\frac{7}{5}\right)$
Find the average value of the function $f(x) = \frac{2x}{x^2 + 1}$ over the interval $[4,7]$.
How do you compute the average value of a function?
You'll need to do a $u$-substitution in here. Make sure you set up the entire expression for average value before you do that.
The expression should be $$ \frac{1}{3}\int_4^7 \frac{2x}{x^2 + 1}\ dx$$
After the substitution, this should be $$ \frac{1}{3} \int_{17}^{50} \frac{1}{u}\ du $$
$\displaystyle \frac{1}{3}\ln\left(\frac{50}{17}\right)$
Let $f(x) = x^2$. Find a value $c$ between 1 and 5 so that $f(c)$ equals the average value of $f$ on $[1,5]$. That is, find the value of $c$ that is guaranteed to exist by the Mean Value Theorem for Integrals.
First, compute the average value of $f$ on this interval.
Once that is done, you want to set $f(c)$ equal to that value and solve for $c$.
You should end up with $ \displaystyle c^2 = \frac{1}{4} \int_1^5 x^2\ dx = \frac{31}{3}$
$\displaystyle \sqrt{\frac{31}{3}} \approx 3.21455$
Find the volume of the solid whose base is the region bounded by the curves $y = x^2 - 1$ and $y = 3 - x^2$, and whose cross sections perpendicular to the x-axis are semicircles with diameter on the $xy$-plane.
Find the average value of the function $f(x) = \sec^2{x}$ on the interval $\left[\frac{\pi}{6}, \frac{\pi}{3} \right]$.