# Math Help - volumes rotate about y = x

1. ## 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

2. 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 $(x', y')= \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ (because that lies on the line y= x and has distance 1 from the origin) and (x,y)= (1, 0) becomes $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$ (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, $x'= \frac{\sqrt{2}}{2}$, $y'= \frac{\sqrt{2}}{2}$, those become (i) $0= \frac{\sqrt{2}}{2}a+ \frac{\sqrt{2}}{2}b$ and (ii) $1= \frac{\sqrt{2}}{2}c+ \frac{\sqrt{2}}{2}d$. Taking x= 1, y= 0, $x'= \frac{\sqrt{2}}{2}$, $y'= -\frac{\sqrt{2}}{2}$, those become (iii) $1= \frac{\sqrt{2}}{2}a- \frac{\sqrt{2}}{2}b$ and (iv) $0= -\frac{\sqrt{2}}{2}c+ \frac{\sqrt{2}}{2}d$.

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

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

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

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

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