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region bounded by $$y=x$$ and $$y=x^2$$

a) find the volume of the solid of revolution formed by revolving R region about the line $x=2$.

b) find the volume of the solid of revolution formed by revolving R region about the line $y=b$ where $b > 1$ is a constant

c) Find any values of a for which the volume of solid in part b is same as volume of solid in part a. Approximate to two decimal places and may use a calculator

I got $\frac \pi2$ for my answer for a but I feel like thats wrong. I integrated with respect to y and found the bounds as 0 to 1 but after calculating I get $\frac \pi2$ for an answer. Please explain in detail how to do these parts and let me know what I did wrong in part a (if its wrong).

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  • $\begingroup$ I think your answer is right, but you might want to check it by integrating with respect to x instead (using the shell method). $\endgroup$
    – user84413
    Commented Sep 7, 2014 at 20:42

2 Answers 2

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Using the shell method,

$V=\displaystyle\int_0^12\pi r(x)h(x)dx=\int_0^12\pi (2-x)(x-x^2)dx=2\pi\int_0^1(2x-3x^2+x^3)\;dx$

$=\displaystyle2\pi\left[x^2-x^3+\frac{x^4}{4}\right]_0^1=2\pi\left(\frac{1}{4}\right)=\frac{\pi}{2}$.

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we have that the region is bounded by $y=x$ and $y=x^2$, finding the intersections

$\left.\begin{matrix}y=x\\y=x^2\end{matrix}\right\} x=x^2\iff x=0\vee x=1$

so, them intercepts on $(0,0)$ and $(1,1)$, then:

a) then the volume of revolution about the line $x=2$, we have that $x=y,x=\sqrt{y}$, then.

for $0\le y\le 1\Rightarrow 2\ge\sqrt{y}\ge y$, wich implies that $(y-2)^2\ge(\sqrt{x}-2)^2$, then we gets that

$$\begin{align}V&=\pi\int_{0}^{1}\left|(y-2)^2-(\sqrt{y}-2)^2\right|dy\\ &=\pi\int_{0}^{1}(y-2)^2-(\sqrt{y}-2)^2dy\\ &=\pi\int_{0}^{1}(y^2-4y+4)-(y-4\sqrt{y}+4)dy\\ &=\pi\int_{0}^{1}y^2-4y+4-y+4\sqrt{y}-4dy\\ &=\pi\int_{0}^{1}y^2-5y+4\sqrt{y}dy\\ &=\pi\left[\frac{y^3}{3}-\frac{5y^2}{2}+\frac{8y\sqrt{y}}{3}\right]_{0}^{1}\\ &=\pi\left(\frac{1}{3}-\frac{5}{2}+\frac{8}{3}\right)\\ &=\pi\frac{2-15+16}{6}\\ &=\frac{\pi}{2}\end{align}$$

b) for this, the solid is revolving over line $y=a$, with $a>1$, then we got:

for $0\le x\le1\Rightarrow a\ge x\ge x^2$, then $(x^2-a)^2\ge(x-a)^2$, then we gets that

$$\begin{align}V&=\pi\int_{0}^{1}\left|(x-a)^2-(x^2-a)\right|dx\\ &=\pi\int-\left[(x-a)^2-(x^2-a)\right]dx\\ &=-\pi\int(x^2-2ax+a^2)-(x^4-2ax^2+a^2)dx\\ &=-\pi\int x^2-2ax+a^2-x^4+2ax^2-a^2dx\\ &=-\pi\int-x^4+(1+2a)x^2-2axdx\\ &=-\pi\left[-\frac{x^5}{5}+\frac{(1+2a)x^3}{3}-ax^2\right]_{0}^{1}\\ &=-\pi\left(-\frac{1}{5}+\frac{1+2a}{3}-a\right)\\ &=-\pi\frac{-3+5(1+2a)-15a}{15}\\ &=-\pi\frac{-3+5+10a-15a}{15}\\ &=-\pi\frac{2-5a}{15}\\ &=\pi\frac{5a-2}{15} \end{align}$$

c) for $c$ we have

$$\frac{\pi}{2}=\pi\frac{5a-2}{15}\\ 15=2(5a-2)\\ 15=10a-4\\ 10a=15+4=19\\ a=\frac{19}{10}=1.9 $$

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  • $\begingroup$ sweet i got it right for part a! :) $\endgroup$
    – MD_90
    Commented Sep 7, 2014 at 21:46
  • $\begingroup$ what about part b and c? $\endgroup$
    – MD_90
    Commented Sep 7, 2014 at 21:47

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