Find the volume of the described solid $ S $.
The base of $ S $ is the same base as in Exercise 56, but cross-sections perpendicular to the x-axis are squares.
Applications of Integration
has given the problem. The cross section on the base corresponding to the coordinates Axe has a length of why therefore, because exports wise one we know why is equivalent to one mine sax. Therefore, if a of exes y squared a vex is one minus x squared because, remember, this is why it's why squared use the foil method which you learned in algebra rock to get one minus two acts plus X squared. So now we have V is the integral from zero to a one of a vexed DX. Well, we just figured out what a of excess. Okay, now that we have this, we know we need to integrate. Use the power rule, increase the export of by one divide by the new experiment, for example, X squared becomes X cubed, divided by three. Plug in. We end up with simply 1/3 because we have one minus one squared plus one cube over +31 minus +101 cube just +11 divided by three is just one over three, which is our solution