I was doing this practice question that if given two functions of y=1/4x and y=x^2, find the volume of the solid around the y-axis.
I am just really confused on what to do - could someone please guide me through it?
Here are my steps so far
Doing 1/4x = x^2, we get x = 0, or x = 0.25.
Thus, continuing to an integral:
$$\int_0^4 pi *((\frac{x}{4})^2 -(x^2)^2) = \int_0^4 pi *((\frac{x^2}{16}) -(x^4)$$
But at this point I get a negative value, so I know I'm doing something wrong.
First off, your integral is missing $dx$, and since you want to find the volume of the solid formed when rotated about the y-axis, you should be looking at the integral with respect to $y$, i.e. an integral $dy$.
This is because you want the cross sections to look about the same and be able to describe them (the cross sections) properly with a function of $y$. (as you move across your solid)
So we write the functions as $x = 4y, x = \sqrt{y}$.
You are correct in finding the points of intersection, but I will turn our focus to the y-coordinate of those points of intersection, and we see they are $y=0,y=\frac{1}{16}$.
So the integral should be phrased as $\displaystyle\int_{0}^{\frac{1}{16}} \pi ((\sqrt{y})^2 - (4y)^2)dy = \int_{0}^{\frac{1}{16}} \pi (y - 16y^2)dy$.