Double Integral of a Circle with Limits of Integration

[LASmith]

Homework Statement

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

Homework Equations

The Attempt at a Solution

using x limits between -1 & 1 followed by the y limits of 0 & [itex]\sqrt{1-x²}[/itex]

[itex]\int[/itex][itex]\int[/itex]y²[itex]\sqrt{1-x²}[/itex].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]

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​

 

[LASmith]

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

[itex]\int[/itex] y³[itex]\sqrt{1-x²}[/itex]/3 .dx

Then substituting the limits [itex]\sqrt{1-x²}[/itex] and 0 for y I obtained

[itex]\int[/itex] (1-x²)²)/3 .dx

Limits between 0 and 1 for this final integral

 

[HallsofIvy]

The set [itex]x^2+ y^2< 1[/itex] 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 [itex]\sqrt{1- x^2}[/itex] (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]

[itex]\int[/itex] (1-x²)²)/3 .dx

= ∫ (1 - 2x² + x⁴)/3 dx …

how did you not get a factor 1/5 after integrating this? ​

 

[LASmith]

= ∫ (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!

… 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?

Just try differentiating (1-x²)³/-18x …

you'd have to use both the chain rule and the product (or quotient) rule, and they don't cancel (why should they?).