# Solids of Revolution

• Feb 13th 2009, 08:11 AM
omgeezy
Solids of Revolution
Really need help, need to check my answers. These are parts of a review so..

#1 Region M is bounded by the x-axis, x=4 and the graph of $LN(x)$

is the interval 1 to 4?

a) find the volume of the solid created by revolving M around y=-4
I got 70.958 as the area

b) find the volume of the solid created by revolving M around y=7
113.964

c) let region M be the base of a solid whose cross sections perpendicular to the x-axis are isosceles right triangles, find the volume of the solid.

d) let region M be the base of a solid whose cross sections perpendicular to the x-axis are circles, find the volume of the solid.
not sure how to do this one either.

#2 Region Q is bounded by the y-axis, the line y=5 and the curve e(x)
Let region R be the base of a solid whose cross sections perpendicular to the x-axis are semi-circles, find the volume of the solid.

Is the interval for the integral 0 to 1.609? And then what do I do? Need help setting it up.

#3 Region T is bounded by the x-axis and the graphs of $\sqrt{x}$ and $-0.2x^2+6$
This one is really confusing me. All I know is the interval 0 to 4.415. I don't understand how you'd solve it on the y-axis. The x-axis was easier to understand.
a) find the volume of the solid created by revolving T around the y-axis
b) around x=-3
c) around x=10
d) let T be the base of a solid whose cross sections perpendicular to the y-axis are squares, find the volume of the solid.
• Feb 13th 2009, 11:42 AM
Reckoner
Quote:

Originally Posted by omgeezy
a) find the volume of the solid created by revolving M around y=-4
I got 70.958 as the area

Using the washer method, we have an outer radius of $R(x) = \ln x - (-4) = \ln x + 4$ and inner radius $r(x) = 0 - (-4) = 4$. You have the right interval, so

$V = \pi\int_1^4\left[(R(x))^2 - (r(x))^2\right]\,dx$

$= \pi\int_1^4\left[(\ln x + 4)^2 - 4^2\right]\,dx$

$= \pi\int_1^4\left[(\ln x)^2 + 8\ln x\right]\,dx$

Is this how you started? I get a slightly different answer.

Quote:

b) find the volume of the solid created by revolving M around y=7
113.964
On $[1,\,4],\;\ln x < 7,$ so your outer radius is $R(x) = 7 - 0 = 7$ and your inner radius $r(x) = 7 - \ln x$. So you have

$V = \pi\int_1^4\left[(R(x))^2 - (r(x))^2\right]\,dx = \pi\int_1^4\left[7^2 - (7 - \ln x)^2\right]\,dx$

Again, I get a different answer.

Quote:

c) let region M be the base of a solid whose cross sections perpendicular to the x-axis are isosceles right triangles, find the volume of the solid.
Which part don't you understand? The region $M$ is the base of the solid; the solid will be "sticking out" of the plane (above the xy-plane, that is). If you cut the solid (with a cut perpendicular to the x-axis), the cross section is an isosceles triangle. This means that the base of each triangle has length $\ln x$.

The formula you need is $V = \int_a^b A(x)\,dx,$ where $A(x)$ is the area of the cross section at $x$.

Quote:

d) let region M be the base of a solid whose cross sections perpendicular to the x-axis are circles, find the volume of the solid.
not sure how to do this one either.
I'm not sure how the solid can have $M$ as a base and still have circular cross sections perpendicular to the x-axis. Semicircles, perhaps?

Quote:

#2 Region Q is bounded by the y-axis, the line y=5 and the curve e(x)
Let region R be the base of a solid whose cross sections perpendicular to the x-axis are semi-circles, find the volume of the solid.

Is the interval for the integral 0 to 1.609? And then what do I do? Need help setting it up.
I assume you mean region $Q$ is the base; you haven't specified $R$. Your interval is correct (approximately). Each semicircle has a base that extends from the top of the region to the bottom of the region. So the diameter of each is $5 - e^x$. Use the formula I gave above.

Quote:

#3 Region T is bounded by the x-axis and the graphs of $\sqrt{x}$ and $-0.2x^2+6$
This one is really confusing me. All I know is the interval 0 to 4.415.
Not quite. $x=4.415$ is roughly the intersection of the two curves, but the region goes beyond that (look at a graph). You want to solve $-0.2x^2 + 6 = 0$ for the upper bound.

I recommend using the washer method and integrating with respect to $y$. In that case, your lower bound is $y = 0,$ and your upper bound is approximately $\sqrt{4.415}\approx2.101\text{.}$ Solve both equations for $x$ (in terms of $y$; you only need the positive root for the second one), and set up your integral.

For the parts involving revolution around horizontal axes, the shell method would be easiest, but you could do it with washers by splitting the interval at the intersection of the curves and using two integrals.