3
$\begingroup$

enter image description here
What is the area of the shared region in the figure above that is bounded by the $x$-axis and the curve with the equation $y=x\sqrt{1-x^2}$?

This is the problem I was given. I assumed the answer was $0$ - the positive and negative areas should cancel each other out. That answer was incorrect. I then thought the answer might be $\frac13$ and they are only asking for the area above the $x$-axis. That was also incorrect. The answer that was given as correct was: $\frac23$. I assume that is because it is the area of the top - which is $\frac13$ and the area of the bottom also $\frac13$ - which makes $\frac23$. But, why is the answer not zero? Doesn't integration count area under the $x$-axis as negative? Does the wording of the question say otherwise?

$\endgroup$
  • 2
    $\begingroup$ Why do you think an area can be negative? That's like saying that the distance between the points $4$ and $3$ is $-1$ beacuse "you go back". The area of a curve is the integral of the absolute value of the function. $\endgroup$ – Alfredo Aug 1 '19 at 20:42
5
$\begingroup$

In your question, it asks for the area of the shaded region, and area is always positive.

For integration, it is taught as the area between the curve and the $x$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $x$-axis, and offset by those below $x$-axis).

Note that Area is absolute, but Net Area is relative

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The integral $\int_{-1}^0y~dx$ is negative, but the area is always non-negative. So the area of the region is $\int_0^1y~dx-\int_{-1}^0y~dx$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The function is odd, the area you look for is $$A=2\int_0^1x\sqrt{1-x^2}dx$$

put $y=x^2$.

then

$$A=\int_0^1\sqrt{1-y}dy$$ $$=\int_0^1(1-y)^{\frac 12}dy$$

$$=\Bigl[ \frac 23(1-y)^{\frac 32}\Bigr]_1^0$$ $$=\frac 23.$$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Let $x=\sin t$.

Thus, we need to get $$2\int\limits_0^1x\sqrt{1-x^2}dx=2\int\limits_0^{\frac{\pi}{2}}\sin{t}\cos^2tdt=\int\limits_0^{\frac{\pi}{2}}\sin2t\cos{t}=$$ $$=\frac{1}{2}\int\limits_0^{\frac{\pi}{2}}(\sin3t+\sin{t})dt=-\frac{1}{6}\cos3t-\frac{1}{2}\cos{t}\big{|}_0^{\frac{\pi}{2}}=\frac{2}{3}.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.