volume of surface of revolution around y axis Can anyone help walk me through this problem style? I have a lot of homework problems like this and I really want to understand how to do these problems.
Find the volume of the solid generated by revolving the given region in the 1st quadrant about the y-axis:
$y=5x^2, x=0, y=245$.
 A: The 2 dimensional area is between the parabola, the y-axis, and the line y=245.  Since you already know the y-bounds, I would suggest you integrate using disks (and hence you integrate over the variable y).
Now your disks are going to be parameterized by y, and are created from horizontal lines extending from the y-axis to the parabola.  One such line would be the radius of a disk when rotated around the y-axis.  That distance is actually the x-coordinate of a point on the parabola, so the disk would have area
$\pi x^2=\pi\frac{y}{5}$ since $5x^2=y$
Now we need to "add" the volume of all such disks, and we do this through integrating.  Since we get a disk for every $0\leq y\leq 245$, we get: $\int_{0}^{245} \pi\frac{y}{5}dy$
A: First, because you are revolving about the y-axis, you need to solve the boundary equation for x.
$$y=5x^2 \rightarrow x=\sqrt{\frac{y}{5}}$$
Before we proceed, you should understand how to find the area of the region that we are revolving. To do this, we integrate the boundary equation as follows.
$$A=\int_0^{245}\sqrt{\frac{y}{5}}dy$$
$$=\frac{2y^{\frac{3}{2}}}{3\sqrt{5}}|_0^{245}$$
$$=\frac{2(245)^{\frac{3}{2}}}{3\sqrt{5}}-\frac{2(0)^{\frac{3}{2}}}{3\sqrt{5}}$$
$$=\frac{3430}{3}\approx1143.33$$
To find the volume of the disk, integrate the square of the boundary equation with respect to $y$ (we are revolving about the y-axis) and multiply by $\pi$.
$$V=\pi\int_0^{245}\sqrt{\frac{y}{5}}^2dy$$
$$=\pi\int_0^{245}\frac{y}{5}dy$$
$$=\frac{\pi y^2}{10}|_0^{245}$$
$$=\frac{\pi (245)^2}{10}-\frac{\pi (0)^2}{10}$$
$$=\frac{12005\pi}{2}\approx18857.41$$
