Compute the area of a given set

Question

Problem

I have the set $D$ defined as:

$$ D=\{(x,y)|x\ge 0, 0 \le y \le 64-81x^2 \} $$

I want to compute the area of this set $A(D)$

Attempt to solve

You could try to solve the area of this set by Riemann sums / integral. First problem is if $x\ge 0$ always and $0 \le y \le 64-81x^2$ $y$ needs to be less than or equal to $64-81x^2$ but when $x$ is positive $64-81x^2$ is always negative?

$0\le y\le64-81x^2$ cannot be true when $x\ge 0$

How do you compute something like this?

Answer 1

Not "always". To have $y\geq0$, you need $64-81x^2\geq0$, that is $$ x^2\leq\frac{64}{81}, $$ which means that $x\leq 8/9$. So your area is $$ \int_0^{8/9}(64-81x^2)\,dx=\frac{64\times 8}9-\frac{81\times(8/9)^3}{3} =64\times8\left(\frac19-\frac1{27}\right)=\frac{1024}{27}. $$

By the way, this has nothing to do with improper integrals.

Answer 2

$f(x)=64-81x^2$ is a concave quadratic function and it is symmetrical about the $y$-axis.

Notice that $$f(0)=64>0.$$

Let's find $x>0$ such that $f(x)=0$.

$$0=64-81x^2$$

$$81x^2=64$$

Hence, it is nonnegative from $0$ to $\frac89$.

$$x=\sqrt{\frac{64}{81}}=\frac89$$

Can you compute the integral now?

Remark: If you are given an empty region, the area would be $0$. However, it is not the case for this question.

Answer 3

$64-81x^2 = 8^2-(9x)^2=0 \iff x = \frac89$ since $x \ge 0$ in $D$.

$$\begin{aligned} A(D) &= \int_0^{8/9} (64-81x^2) dx \\ &= [64x - 27x^3]_0^{8/9} \\ &= 64 \cdot \frac89 - 27\left(\frac89\right)^3 \\ &= \frac{512}{9} - 27 \cdot \frac{512}{729} \\ &= \frac{1024}{27} \end{aligned}$$

Answer 4

A bit of coordinate geometry.

$x\ge 0$, $y\ge 0$ implies

we are looking at the $1$st quadrant only.

Consider the parabola:

$y = 64 - 81x^2$,

vertex $(0,64)$ , opening downward .

Summing up :

You want to find the area bounded by $X-,Y-$axes, and by $y = 64-81x^2.$

Set $y=0$ to find the point if intersection with the $X-$axis.

$64-81x^2=0; $

$x_1=\sqrt{\dfrac{64}{81}}=8/9.$

(Since we are confined to the $1$st quadrant only the positive root is of interest.)

The area: $A = \displaystyle \int_{0}^{8/9}(64-81x^2)dx.$

Remains to carry out the integration.