# Thread: Using Definite Integral, Applications to Geometry

1. ## Using Definite Integral, Applications to Geometry

A pie dish is 9inches across the top, 7 inches across the bottom, and 3 inches keep. Compute the volume of this dish.

2. ## Solutions!

SO to start this bad boy off, you should start by drawing a picture. Since i am not entirely sure if we can draw a picture i will do my best to describe this.

First put the pie dish on the origin. The x-axis should be through the middle of the dish. The bottom of the dish should be sitting on the y-axis. It should look like it is on it's side. The total height should be 4.5, and the total length should be 3. It should open to the right. Only draw the parts that are in quadrant 1. The actual dish will be in both quadrant 1 and 4, but we are going to get rid of half of it. The reason is because we rotate around the axis 360 degrees, if we rotated the part in 4, we would end up with double the volume.

Now we need to find the equation of the line. We need a slope, and a point to do that. (Ok there are many ways to do it but i like this one.) We actually have 2 points,
$(3,4.5)$ $(0,3)$
To find the slope we do
$m= rise/run = (4.5-3)/3 = .5$.

Now we put into formula
$y-y_1 = m(x-x_1)$
$y=.5x+3$

Now that we have our line, we need to do some integrals, and rotating.
Now we can use the disc method.
$V=\pi\int^a_bR^2 dx$
The "R" in this equation is the distance from your function to the axis of rotation, in this case it is .5x+3.
The limits are from 0 to 3 because that is how deep your dish is.
$V=\pi\int^3_0(.5x+3)^2 dx$
$V=\pi\int^3_0(.25x^2+3x+9)dx$
$V=\pi[\frac{1}{12}x^3+\frac{3}{2}x^2+9x]^3_0dx$
$V=\frac{171}{4}\pi$