area and volume in 2 different ways

logistic_guy

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here is the question

Consider the first-quadrant region R\displaystyle \mathcal{R} enclosed by the curves y=x2\displaystyle y = x^2 and y=x\displaystyle y = \sqrt{x}.

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


my attemb
x2=x\displaystyle x^2 = \sqrt{x}
x2=x\displaystyle \sqrt{x^2} = \sqrt{\sqrt{x}}
x=x\displaystyle x = \sqrt{\sqrt{x}}
xx=0\displaystyle x - \sqrt{\sqrt{x}} = 0
i think 1 give zero
11=0\displaystyle 1 - \sqrt{\sqrt{1}} = 0
so the region is in the intevral 0x1\displaystyle 0 \leq x \leq 1
is it correct?
 
Draw the graphs. The points of intersection are clearly (0, 0) and (1, 1). So, yes the interval is 0x1\displaystyle 0\leq x\leq 1.
 
Draw the graphs. The points of intersection are clearly (0, 0) and (1, 1). So, yes the interval is 0x1\displaystyle 0\leq x\leq 1.
thank

area.png

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

r1=(t,t2)\displaystyle \bold{r}_1 = (t,t^2)

r2=(t,t)\displaystyle \bold{r}_2 = (t,\sqrt{t})
 
If I were you, I'd find a calculus textbook, such as this:

As for finding the intersections algebraically, it's easier to either square both sides of x2=xx^2=\sqrt{x}, then get 0 on one side and factor; or just factor out x\sqrt{x} from both terms of x2x=0x^2-\sqrt{x}=0. Taking square roots just makes it harder to see.
 
thank Dr.

If I were you, I'd find a calculus textbook, such as this:
i'll read it

As for finding the intersections algebraically, it's easier to either square both sides of x2=xx^2=\sqrt{x}, then get 0 on one side and factor; or just factor out x\sqrt{x} from both terms of x2x=0x^2-\sqrt{x}=0. Taking square roots just makes it harder to see.
after i post my attemb i realize i can square

x2=x\displaystyle x^2 = \sqrt{x}
x4=x\displaystyle x^4 = x
x4x=0\displaystyle x^4 - x = 0
x(x31)=0\displaystyle x(x^3 - 1) = 0

i don't never thinkg of factor x\displaystyle \sqrt{x}
x2=x\displaystyle x^2 = \sqrt{x}
x2x=0\displaystyle x^2 - \sqrt{x} = 0
xx1.5x=0\displaystyle \sqrt{x}x^{1.5} - \sqrt{x} = 0
x(x1.51)=0\displaystyle \sqrt{x}(x^{1.5} - 1) = 0
 
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