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Set up a definite integral that represents the volume obtained by rotating the region between the curve $y^{2}=x$ and $y^{2}=2(x-1)$ about the lines below:

(a) Rotate about the y-axis

(b) Rotate about the line $y=3$

The question itself is not hard but I'm having some trouble deciding as to how I should set up the integral. Any guidance?

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  • $\begingroup$ After some thought I get: $\int\limits_{-\sqrt{2}}^{\sqrt{2}}\pi(x^2-(2(x-1))^2) dx$ is that right? $\endgroup$ – Future Math person Jul 3 '16 at 4:15
  • $\begingroup$ For part a, you want to use the shell method if you integrate with respect to x, or the disc method if you integrate with respect to y. $\endgroup$ – user84413 Jul 3 '16 at 4:24
  • $\begingroup$ Yep I am using the disc method. $\endgroup$ – Future Math person Jul 3 '16 at 4:32
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When rotated around the y axis the integral is $V=\int \pi x^2 dy$.

in your case it would be $$\int \pi (x_2^2-x_1^2)dy$$ where $x_2=f(y),x_1=g(y)$ .

first of all you take a strip of width $dy$ around the $y$ axis and find the volume of that infinitesimal space generated.

Intuitively it becomes clear that the volume will be the area of the disk multiplied by the element $dy$ to get volume of revolution. $$\int^{\sqrt2}_{-\sqrt2} \pi(y^4-(y^2/2+1)^2)dy$$

b) rotating about the $y=3$ $$V=\pi\int^{2}_{0} ((3-\sqrt x)^2-(3-\sqrt{2x-1})^2)dx$$

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  • $\begingroup$ Okay so after thinking I get: $\int\limits_{0}^{2} \pi(x-2(x-1))dx$ Is that right? $\endgroup$ – Future Math person Jul 3 '16 at 4:49
  • $\begingroup$ i edited above @SubhashisChakraborty $\endgroup$ – Nebo Alex Jul 3 '16 at 4:55
  • $\begingroup$ I see it but should it not be $x=f(y)$ and not $f(x)$ ? Otherwise we have a function of x inside an integral with the differential being dy so the x's would be treated as a constant. $\endgroup$ – Future Math person Jul 3 '16 at 4:58
  • $\begingroup$ yes you are right $\endgroup$ – Nebo Alex Jul 3 '16 at 5:08
  • $\begingroup$ Now I get: $\int\limits\limits_{-\sqrt{2}}^{\sqrt{2}} \pi(y^4-\ (\frac{ y^2+2 }{ 2 })^{2})dy$ $\endgroup$ – Future Math person Jul 3 '16 at 5:13
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$\textbf{a)}$ Using the disc method and symmetry, we have

$\hspace{.2 in}\displaystyle V=2\int_0^{\sqrt{2}}\pi((R(y))^2-(r(y))^2)dy=\color{blue}{2\int_0^{\sqrt{2}}\pi\left(\left(\frac{y^2}{2}+1\right)^2-(y^2)^2\right)dy}$


$\;\;\;$Alternate answer: (shell method)

$\hspace{.2 in}\displaystyle V=\color{green}{\int_0^{2}2\pi x\big(2\sqrt{x}\big)dx-\int_1^{2}2\pi x\big(2\sqrt{2(x-1)}\big)dx}$

$\textbf{b)}$ Using the shell method,

$\hspace{.2 in}\displaystyle V=\int_{-\sqrt{2}}^{\sqrt{2}}2\pi r(y) h(y)dy=\color{blue}{\int_{-\sqrt{2}}^{\sqrt{2}}2\pi (3-y)\left(\left(\frac{y^2}{2}+1\right)-y^2\right)dy}$


$\;\;\;$Alternate answer: (disc method)

$\hspace{.2 in}\displaystyle V=\color{green}{\int_0^{2}\pi\left((3+\sqrt{x})^2-(3-\sqrt{x})^2\right)dx-\int_1^{2}\pi\left(\left(3+\sqrt{2(x-1)}\right)^2-\left(3-\sqrt{2(x-1)}\right)^2\right)dx}$

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