Find the centroid (x¯,y¯)(x¯,y¯) of the region bounded by:

y=5x^2+6x
y=0
x=0
x=6
I know the integral (think, at least) is from 0 to 6 and between the first and second equation given, but I missed this lecture and have no idea how to find the center of mass

To find the centroid of the given region, we need to find the x-coordinate (x¯) and y-coordinate (y¯) of the centroid separately using the formulas for centroid calculations.

The centroid of a region in the xy-plane can be found by performing the following steps:

1. Find the area of the region: This can be done by calculating the definite integral of the function that bounds the region from the lower bound of x=0 to the upper bound of x=6. In this case, the region is bounded by the curve y=5x^2+6x and the x-axis, so the area can be found by integrating this function with respect to x over the interval [0, 6].

Area (A) = ∫[0, 6] (5x^2 + 6x) dx

To calculate the integral, you need to find the antiderivative of the given function and evaluate it at the upper and lower bounds:

A = [(5/3)x^3 + 3x^2] evaluated from 0 to 6
= [(5/3)(6^3) + 3(6^2)] - [(5/3)(0^3) + 3(0^2)]
= (5/3)(216) + 3(36)
= 360 + 108
= 468

So, the area of the region is 468 square units.

2. Find the x-coordinate of the centroid (x¯): The x-coordinate of the centroid can be determined by using the formula as follows:

x¯ = (1/A) * ∫[0, 6] (x * (5x^2 + 6x)) dx

Here, A represents the area of the region we calculated in step 1.

Simplifying the above expression and calculating the integral, we get:

x¯ = (1/468) * ∫[0, 6] (5x^3 + 6x^2) dx

x¯ = (1/468) * [(5/4)x^4 + (6/3)x^3] evaluated from 0 to 6

x¯ = (1/468) * [(5/4)(6^4) + (6/3)(6^3)] - [(5/4)(0^4) + (6/3)(0^3)]

x¯ = (1/468) * [(5/4)(1296) + (6/3)(216)] - 0

x¯ = (1/468) * [(5/4)(1296) + (6/3)(216)]

x¯ = (1/468) * (1620 + 432)

x¯ = 2052/468

x¯ = 4.385

So, the x-coordinate of the centroid is approximately 4.385.

3. Find the y-coordinate of the centroid (y¯): The y-coordinate of the centroid can be determined using the formula as follows:

y¯ = (1/(2A)) * ∫[0, 6] (((5x^2 + 6x)^2)/2) dx

Simplifying the above expression and calculating the integral, we get:

y¯ = (1/(2*468)) * ∫[0, 6] (5x^2 + 6x)^2 dx

y¯ = (1/936) * ∫[0, 6] (25x^4 + 60x^3 + 36x^2) dx

Performing the integral, we get:

y¯ = (1/936) * [(5/5)x^5 + (60/4)x^4 + (36/3)x^3] evaluated from 0 to 6

y¯ = (1/936) * [(1/5)(6^5) + (1/4)(60)(6^4) + (1/3)(36)(6^3)] - 0

y¯ = (1/936) * [(1/5)(7776) + (1/4)(60)(1296) + (1/3)(36)(216)]

y¯ = (1/936) * (1555.2 + 1944 + 432)

y¯ = 3931.2/936

y¯ = 4.20

So, the y-coordinate of the centroid is approximately 4.20.

Therefore, the centroid (x¯, y¯) of the given region bounded by y=5x^2+6x, y=0, x=0, x=6 is approximately (4.385, 4.20).

google the topic and you will find many discussions and examples.

For a region of uniform density, the center of mass is the same as the centroid.