area and volume in 2 different ways

[logistic_guy]

here is the question

Consider the first-quadrant region \(\displaystyle \mathcal{R}\) enclosed by the curves \(\displaystyle y = x^2\) and \(\displaystyle y = \sqrt{x}\).

(a) Find the area of \(\displaystyle \mathcal{R}\).
(b) Find the volume of the solid that is obtained when \(\displaystyle \mathcal{R}\) is revolved around the \(\displaystyle x\)-axis.
Check your answer by redoing the problem using a different method.
(c) The centroid of \(\displaystyle \mathcal{R}\) is (of course) somewhere on the line \(\displaystyle y = x\). Where?


my attemb
\(\displaystyle x^2 = \sqrt{x}\)
\(\displaystyle \sqrt{x^2} = \sqrt{\sqrt{x}}\)
\(\displaystyle x = \sqrt{\sqrt{x}}\)
\(\displaystyle x - \sqrt{\sqrt{x}} = 0\)
i think 1 give zero
\(\displaystyle 1 - \sqrt{\sqrt{1}} = 0\)
so the region is in the intevral \(\displaystyle 0 \leq x \leq 1\)
is it correct?

 

[Harry_the_cat]

Draw the graphs. The points of intersection are clearly (0, 0) and (1, 1). So, yes the interval is \(\displaystyle 0\leq x\leq 1\).

 

[logistic_guy]

Draw the graphs. The points of intersection are clearly (0, 0) and (1, 1). So, yes the interval is \(\displaystyle 0\leq x\leq 1\).

thank



i know this question is solve by calculus. i study calculus long time and i forgot
i'm advance engineering, so i'm very good in differential geometry. i'm think of transforming the functions to curves

\(\displaystyle \bold{r}_1 = (t,t^2)\)

\(\displaystyle \bold{r}_2 = (t,\sqrt{t})\)

 

[Dr.Peterson]

If I were you, I'd find a calculus textbook, such as this:

Calculus I - Area Between Curves

In this section we’ll take a look at one of the main applications of definite integrals in this chapter. We will determine the area of the region bounded by two curves.

tutorial.math.lamar.edu

As for finding the intersections algebraically, it's easier to either square both sides of [imath]x^2=\sqrt{x}[/imath], then get 0 on one side and factor; or just factor out [imath]\sqrt{x}[/imath] from both terms of [imath]x^2-\sqrt{x}=0[/imath]. Taking square roots just makes it harder to see.

 

[logistic_guy]

thank Dr.

If I were you, I'd find a calculus textbook, such as this:

Calculus I - Area Between Curves

In this section we’ll take a look at one of the main applications of definite integrals in this chapter. We will determine the area of the region bounded by two curves.

tutorial.math.lamar.edu

i'll read it

As for finding the intersections algebraically, it's easier to either square both sides of [imath]x^2=\sqrt{x}[/imath], then get 0 on one side and factor; or just factor out [imath]\sqrt{x}[/imath] from both terms of [imath]x^2-\sqrt{x}=0[/imath]. Taking square roots just makes it harder to see.

after i post my attemb i realize i can square

\(\displaystyle x^2 = \sqrt{x}\)
\(\displaystyle x^4 = x\)
\(\displaystyle x^4 - x = 0\)
\(\displaystyle x(x^3 - 1) = 0\)

i don't never thinkg of factor \(\displaystyle \sqrt{x}\)
\(\displaystyle x^2 = \sqrt{x}\)
\(\displaystyle x^2 - \sqrt{x} = 0\)
\(\displaystyle \sqrt{x}x^{1.5} - \sqrt{x} = 0\)
\(\displaystyle \sqrt{x}(x^{1.5} - 1) = 0\)