Question

How do you do these types of questions?

Answer 1

Answer:

V = 11π/30, 'Sketch the region' = Second Attachment, 'Sketch the solid, and a typical dish or washer' = First Attachment

Step-by-step explanation:

Let's start by determining the intersection points of y = x² and x = y². The procedure would be as follows:

x = y²,

x = (x²)² = x⁴,

0 = x⁴ - x,

0 = x(x³ - 1),

x = 0, and x³ - 1 = 0

Solution(s): x = 0, and x = 1

If we want to create a 'three dimensional graph' then we would have to graph the functions y = x² and x = y² between their intersection points at x = 0, and x = 1. Then rotate the region about the line 'y = 1.' Our graph will be as demonstrated in the first attachment.

This cross section is a washer, as it is present with a hole. The top bound of the region is x = y², or y = √x. The bottom bound of the region is y = x². The large radius would be R = 1 - x², and the small radius would be r = 1 - x^1/2. The volume would then be:

$V\:=\:\pi \int _0^1\:\left[\left(1-x^2\right)^2-\left(1-x^{\frac{1}{2}}\right)^2\right]dx,\\=> \pi (\frac{1}{5}-\frac{2}{3}-\frac{1}{2}+\frac{4}{3})\\\\=> \pi \frac{11}{30}\\\\=>\frac{11\pi }{30}$

As you can see your volume = 11π/30. Your 'two dimensional' graph would just be a sketch of the given function's parabola(s). Take a look at the second attachment for your two dimensional graph.