Finding the volume of revolution using the method of shells

Question

I'm trying to find the volume of the solid generated by revolving the region bounded by $y=x^2$ and $y=6x+7$ about $x$-axis using the shell method. I applied the method and I got $15864/5$ multiplied by $\pi$ but it's not correct.

Details: I integrated $$\int_{1}^{49}y(\sqrt{y}-\frac{y-7}{6})dy$$

Answer 1

For revolving about the $x$-axis, use washers. Then $V=\pi\int_{-1}^7 [(6x+7)^2-(x^2)^2]dx$

Answer 2

If you want to use the shell method, then you would need

$\displaystyle\int_0^{1}2\pi y\left(\sqrt{y}-(-\sqrt{y})\right) dy +\int_{1}^{49}2\pi y\left(\sqrt{y}-\frac{y-7}{6}\right) dy$;

but using the disc method as indicated in the other answer is easier.