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Find the centroid of the thin plate bounded by the graphs of f(x) = x^{2} and g(x) = x + 6.

Question:

Find the centroid of the thin plate bounded by the graphs of {eq}f(x) = x^{2} {/eq} and {eq}g(x) = x + 6 {/eq}.

Centroid of a Region Bounded by Two Functions

In getting the centroid of a region, integration method can be applied. If a region {eq}R {/eq} lies between two functions or curves defined by {eq}y = f(x) {/eq} and {eq}y = g(x) {/eq} and {eq}f(x)\geq g(x) {/eq}, then the centroid of the region is given by the coordinates {eq}(\bar{x},\bar {y}) {/eq},

where

$$\begin{align*} \bar{x}&=\frac{1}{A}\int_{a}^{b}x[f(x)-g(x)]dx\\ \bar{y}&=\frac{1}{2A}\int_{a}^{b}([f(x)]^{2}-[g(x)]^{2})dx \end{align*} $$

{eq}A {/eq} is the area of the region bounded by the two curves {eq}f(x) {/eq} and {eq}g(x) {/eq}. Its value is given by the formula

$$A=\int_{a}^{b}[f(x)-g(x)]dx $$

Answer and Explanation:


The first step is to find the value of the area of the region {eq}A {/eq} by applying the formula to the functions.

$$A=\int_{a}^{b}[f(x)-g(x)]dx $$

We have the functions {eq}y =x+6 {/eq} and {eq}y = x^2 {/eq}. We must first evaluate the limits of integration by equating them

$$\begin{align*} x+6&=x^2\\ x^2-x-6&=0\\ (x-3)(x+2)&=0\\ x&= -2,3 -2 \leq &x\leq 3 \end{align*} $$

Before we evaluate the area, we must first make sure which function is above the region and which one is below. To do this, we can use an arbitrary value within the function and substitute it into each function. The function that gives the higher value is the function above the region. For the value within the limits, we can use {eq}0 {/eq}.

$$\begin{align*} f(0) &= 0^2 = 0\\ g(0) &= 0 + 6 = 6 \end{align*} $$

Since {eq}g(x) > f(x) {/eq}, we can say that {eq}g(x) {/eq} and {eq}f(x) {/eq} is below the region. This means that in the formula for the area, we are subtracting {eq}f(x) {/eq} from {eq}g(x) {/eq}. We can now evaluate the area.

$$\begin{align*} A&=\int_{-2}^{3}[x+6-x^2]dx\\ A&=\left | \frac{x^2}{2}+6x-\frac{x^3}{3} \right |_{-2}^{3} \\ A&=\frac{5}{2}+30-\frac{35}{3} \\ A&=\frac{125}{6} \end{align*} $$


We can now find the values of the coordinates of the centroid of the region using the given formula. We will start with the x-coordinate.

$$\begin{align*} \bar{x}&=\frac{1}{A}\int_{a}^{b}x[f(x)-g(x)]dx\\ \bar{x}&=\frac{6}{125}\int_{-2}^{3}x(x+6-x^2)dx\\ \bar{x}&=\frac{6}{125}\int_{-2}^{3}(x^2+6x-x^3)dx\\ \bar{x}&=\frac{6}{125}\left | \frac{x^3}{3}+3x^2-\frac{x^4}{4} \right |_{-2}^{3}\\ \bar{x}&=\frac{6}{125} \left(\frac{35}{3}+15-\frac{65}{4} \right)\\ \bar{x}&=\frac{6}{125}\times \frac{125}{12}\\ \bar{x}&=\frac{1}{2} \end{align*} $$


Next, we will evaluate the value of the y-coordinate of the centroid of the region.

$$\begin{align*} \bar{y}&=\frac{1}{2A}\int_{a}^{b}([f(x)]^{2}-[g(x)]^{2})dx\\ \bar{y}&=\frac{3}{125}\int_{-2}^{3}[ (x+6)^2-x^4 ]dx\\ \bar{y}&=\frac{3}{125}\int_{-2}^{3}(x^2+36+12x-x^4)dx\\ \bar{y}&=\frac{3}{125} \left(36x+\frac{x^3}{3}+6x^2-\frac{x^5}{5} \right) \biggr|_{-2}^{3}\\ \bar{y}&=4 \end{align*} $$


Hence, the centroid of the region bounded by the curve is {eq}\left(\frac{1}{2}, 4 \right) {/eq}.


Learn more about this topic:

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