Integral Applications story problem

[medicalphysicsguy]

Greetings,

This is a story problem in my Swokowski Calculus book. I suspect the answer at the back of the book is wrong because I find it hard to believe the answer doesn't have pi in it. I am self-tutoring so any help is greatly appreciated.

"A log having the shape of a right circular cylinder of radius a is lying on its side. A wedge is removed from the log by making a vertical cut and another cut at the angle of 45 degrees, both cuts intersecting at the center of the log. Find the volume of the wedge."

I get: 1/2*pi*a^3 - a^3 .

Book says: 2/3*a^3.

Is it really possible a wedge cut from a cylinder does not have pi in the answer??

My reasoning is, the x length of the wedge is a due to the 45 degree cut. For a given partition approaching size 0, the volume will be the half-circle representing that slice of the cylinder times dx, minus the square representing that slice of the wedge times dx. Therefore, you integrate from 0 to a the area of the half-circle 1/2*pi*a^2dx minus the area of the square 2axdx.

Can anyone help, thank you.

 

[tkhunny]

The cross section of the wedge has no circular portion. Does your interpretaton include a circular portion?

 

[galactus]

Picture right triangles going along the length of the wedge with height h and length x.

The wedge is cut with angle \(\displaystyle \theta\)

Thus, we have \(\displaystyle tan\theta = \frac{h}{x}, \;\ h=xtan\theta\)

The area of each triangle is \(\displaystyle A(y)=\frac{1}{2}hx=\frac{1}{2}x^{2}tan\theta\)

\(\displaystyle =\frac{1}{2}(a^{2}-y^{2})tan\theta\).

because \(\displaystyle x^{2}=a^{2}-y^{2}\).

\(\displaystyle V=\displaystyle \frac{1}{2}tan\theta \int_{-a}^{a}(a^{2}-y^{2})dy\)

\(\displaystyle =\displaystyle tan\theta \int_{0}^{a}(a^{2}-y^{2})dy=\frac{2}{3}a^{3}tan\theta\)

Since \(\displaystyle tan(\frac{\pi}{4})=1\), we get \(\displaystyle \frac{2}{3}a^{3}\)

No Pi

 

[medicalphysicsguy]

Thanks

Many thinks to you both. I found another textbook that uses the exact same method you describe. My reasoning was way off. I don't think my book chapter got me ready for this problem. There was no trig in the chapter, and we really haven't tackled 3-D formally yet, just some tricks for solids that involve rotating a shape around an axis, which I don't think was sufficient for this problem.