Problem 69 Hard Difficulty
A hole of radius $ r $ is bored through the middle of a cylinder of radius $ R > r $ at right angles to the axis of the cylinder. Set up, but do not evaluate, an integral for the volume cut out.
Answer
$V=8 \int_{0}^{r} \sqrt{r^{2}-z^{2}} \sqrt{R^{2}-2^{2}} d z$
More Answers
Topics
Applications of Integration
Discussion
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Anna Marie V.
Campbell University
Kristen K.
University of Michigan - Ann Arbor
Samuel H.
University of Nottingham
Watch More Solved Questions in Chapter 6
Video Transcript
we know the general Methodist integrate from zero to our and the multiplied by two. Because again we have the cemetery property, which means multiplying by two will gust the correct solution rather than having negative bounds. In this context, which means at Z based on the circle with the radius we end up with r squared is why squared plus disease squared. Consequently, capital R squared is X squared plus c squared. They're for one whole cross section. Remember, How are multiplying intro by two into running from zero r is gonna be four times squirt of r squared minus C squared times square, root of capital R squared minus Z squared and then times D Z And then it just said to set up the integral and not to multiply it. But the one thing we can do, though, to make this more sympathize toe make the four times to eight That just makes it a little cleaner. But overall, we don't have to actually solve the integral for this problem. All they've asked us to do is write it out because it's kind of a complicated into girl to solve anyway.
Topics
Applications of Integration
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Anna Marie V.
Campbell University
Kristen K.
University of Michigan - Ann Arbor
Samuel H.
University of Nottingham
