# volumes rotate about y = x

• Apr 27th 2010, 03:00 AM
differentiate
volumes rotate about y = x
i don't know how to start this problem. Instead of rotating about a horizontal and vertical line, it requires you to rotate about the line y =x

determine the volume bounded by the curve when y = x^4 and y =x between x = 0 and x = 1 are rotated about the line y =x

Thanks
• Apr 27th 2010, 07:12 AM
HallsofIvy
That's going to be a difficult problem but I think I would be inclined to rotate the coordinate system so as to make the line y= x correspond to the new y-axis. That is, we must rotate so that (x, y)= (0, 1) becomes $\displaystyle (x', y')= \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ (because that lies on the line y= x and has distance 1 from the origin) and (x,y)= (1, 0) becomes $\displaystyle \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$ (because that lies on y= -x, the line perpedicular to y= x, in the fourth quadrand and has distance 1 from the origin).

Since a rotation is linear, we must have x= ax'+ by' and y= cx'+ dy'. Taking x= 0, y= 1, $\displaystyle x'= \frac{\sqrt{2}}{2}$, $\displaystyle y'= \frac{\sqrt{2}}{2}$, those become (i) $\displaystyle 0= \frac{\sqrt{2}}{2}a+ \frac{\sqrt{2}}{2}b$ and (ii)$\displaystyle 1= \frac{\sqrt{2}}{2}c+ \frac{\sqrt{2}}{2}d$. Taking x= 1, y= 0, $\displaystyle x'= \frac{\sqrt{2}}{2}$, $\displaystyle y'= -\frac{\sqrt{2}}{2}$, those become (iii) $\displaystyle 1= \frac{\sqrt{2}}{2}a- \frac{\sqrt{2}}{2}b$ and (iv)$\displaystyle 0= -\frac{\sqrt{2}}{2}c+ \frac{\sqrt{2}}{2}d$.

Adding (i) and (iii) gives $\displaystyle 1= \sqrt{2}a$ so $\displaystyle a= \frac{\sqrt{2}}{2}$. (i) is the same as b= -a so $\displaystyle b= -\frac{\sqrt{2}}{2}$.

Adding (ii) and (iv) gives $\displaystyle 1= \sqrt{2}d$ so $\displaystyle d= \frac{\sqrt{2}}{2}$. (iv) is the same as d= c so $\displaystyle c= \frac{\sqrt{2}}{2}$.

That is, $\displaystyle x= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'$ and $\displaystyle y= \frac{\sqrt{2}}{2}x+ \frac{\sqrt{2}}{2}y$.

Now, the region is bounded by $\displaystyle y= x^4$ and y= x itself. Obviously, y= x becomes just y'= 0. $\displaystyle y= x^4$ becomes $\displaystyle \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y= (\frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y')^4$.

I will leave it to you to multiply that out. You are now rotating that around the x' axis from x'= 0 to $\displaystyle \sqrt{2}$.