The base of S is a circular disk with radius 5r. Parallel crosssections perpendicular to the base are squares.
1 Answer

Ok. So you will need an integral for sure ok?
Your limits will be from 5r to 5r. you can draw the bottom disk. Its a circle around the origin. Its highest x point is 5 r and smallest is 5r right?
Now you need some equation to evaluate. You need to know that volumue is usually an integral of some area.
What shape do they give you? A square. The cross sections are squares.
But how do u find the area of each square?
If you drew the circle and you drew some line thru it. Thats the bottom of the square.
In other words the height of the bottom disk is the lenght of the side of a square.
Its really hard to explain in words without oictures.
But think of this
The equation of your circle is sqrt(25r^2 x^2). This gives you onky half the heightZ. So multiply by two.
And you have your side of the square
Which is
2sqrt(25r^2 – x^2)
How do you find area a square? Its side square.
And you are done!
All you need is to evaulate
Integral (from 5r to 5r) [ 2sqrt(25r^2 – x^2) ] ^2 dx
Generally jts the integral of the area of tr cross section you are doing.
Always
So i can evaluate for you
I am sure u can do integrals
the answer should be
8( 125r^3 – (125/3)r^3)
Or in super simplied form
(2000/3)r^3