Volume of a solid with base bounded by a parabola and a straight line

Question

The base of a solid is the region bounded by $y=0,25x^2$ and $y=3-x$. Cross sections perpendicular to the $y$ axis are rectangles with base in the XY plane and its height is twice its length. Find the area of the rectangle and then the volume of the solid.

Well, first I found the intersection points of the two curves, they are $A=(-6;9)$ and $B=(2;1)$. Now I want to find the length of the rectangle, and here's where I have some problems. Because from $y=1$ to $y=9$ the side of the rectangle goes from the parabola to the straight line, but from $y=0$ to $y=1$ it goes from the parabola to the parabola. So probably I have to compute two calculations, but I don't know what two calculations are.

Answer 1

The trick here is to split the interval into $[0,1)$ and $(1,9]$ and integrate separately.

First for $[0,1)$, the length of the rectangle is $2\sqrt{4y}$. So height is $4\sqrt{4y}$. So you can integrate $$\int_0^18(\sqrt{4y})^2 dy=\int_0^132y^2 dy$$

Then for $(1,9]$, the length of the rectangle is $3-y+\sqrt{2y}$. So height is $2(3-y+\sqrt{2y})$. So you can integrate $$\int_0^12(3-y+\sqrt{2y})^2 dy$$.

Then add the two results together to get the total volume.

Answer 2

Like you thought, from $y = 1$ to $y = 9$, the integral is between the line and the parabola, $$\int_1^9 2(3-y-(-2\sqrt{y}))^2dx$$

From $y = 0$ to $y = 1$, it is between the left and right side of the parabola $$\int_0^1 2((2\sqrt{y})-(-2\sqrt{y}))^2dy$$

Add these two together to get the complete volume.

To get the area, it is a simple integration from $x = -6$ to $x = 2$

$$\int_{-6}^{2} (3-x)-(0.25x^2)dx$$