# Double integral circle limits

## Homework Statement

Evaluate
f(x,y)=y2$\sqrt{1-x2}$
over the region
x2+y2< 1

## The Attempt at a Solution

using x limits between -1 & 1 followed by the y limits of 0 & $\sqrt{1-x2}$

$\int$$\int$y2$\sqrt{1-x2}$.dy.dx

Evaluating this and multiplying be 2 to get the whole circle I get 0 as the 1 and -1 limits both give zero as the answer. Therefore I used the x limits of 0 & 1 and multiplied by 4 instead, this gave an answer of 4/9, however I have been given the answer of 32/45, and I have no idea how this could possibly be wrong.

tiny-tim
Homework Helper
Hi LASmith! … Therefore I used the x limits of 0 & 1 and multiplied by 4 instead, this gave an answer of 4/9 …

Your intermediate integrand must be wrong. Show us what you had inside the ∫ when it was just ∫ … dx​

Show us what you had inside the ∫ when it was just ∫ … dx​

$\int$ y3$\sqrt{1-x2}$/3 .dx

Then substituting the limits $\sqrt{1-x2}$ and 0 for y I obtained

$\int$ (1-x2)2)/3 .dx

Limits between 0 and 1 for this final integral

HallsofIvy
Homework Helper
The set $x^2+ y^2< 1$ is the interior of a circle with center at (0, 0) and radius 1. That means that a diameter, on the x-axis, goes from -1 to 1, and on the y-axis, from -1 to 1. By taking y from 0 (the x-axis) up to $\sqrt{1- x^2}$ (the circle) and x from -1 to 1, you are only integrating over 1/2 of the circle.

Your y needs to go from the point where the vertical line crosses the circle below the x-axis to the point where it crosses the circle above the x-axis.

tiny-tim
Homework Helper
$\int$ (1-x2)2)/3 .dx

= ∫ (1 - 2x2 + x4)/3 dx …

how did you not get a factor 1/5 after integrating this? = ∫ (1 - 2x2 + x4)/3 dx …

how did you not get a factor 1/5 after integrating this? Expanding the brackets and then integrating I obtained the correct answer, so thanks for that.

However, looking back I had no idea how I did not get a factor of 1/5, I did not expand out the brackets, and just used the chain rule to obtain
(1-x2)3/-18x
Then putting in the limits 0 and 1 gets the answer 0-0, out of curiosity how does this method not work?

tiny-tim
Hi LASmith! 