Question

Double Integrals

Answer 1

Im having difficulty finding the correct bound to use for my integrals and am unsure if I need to convert to polar.

Answer 2

Yeah your region is on a disc of radius 1. You can either do it in rectangular coordinates as something like:

\[\int_{x=-1}^{x=1} \int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} f(x,y) dy dx\]
Or convert to polar and do:
\[\int_0^{2\pi} \int_0^1 f(r,\theta) r dr d \theta\]

Answer 3

Or you can note by symmetry that you are integrating an entirely odd function over an even domain so to speak; which makes it 0.

Where the integral is +x on the right of the circle will exactly cancel with where it's -x on the left of the circle. Similar circumstance for y^3.

Answer 4

so \(-\sqrt{1-x^2}\) accounts for the bottom of the circle up to the x-axis and \(\sqrt{1-x^2}\) accounts for the x axis to the top of the circle? How do I know whether to integrate with respect to x first or y first? Thanks.

Answer 5

The order of integration is up to you. Currently, you're given a double integral over a general region,
\[I=\iint_\Omega (x+7y^3)\,\mathrm dx\,\mathrm dy\]which doesn't tell you how the function gets integrated over the region. The order of the differential elements here is arbitrary.

Once you determine the order you'd like to use (typically in the direction of one variable first, then the other) you can work on figuring out the constraints.

So if you choose to integrate with respect to \(x\) then \(y\), you would write
\[I=\int_{-1}^1\int_{-\sqrt{1-y^2}}^\sqrt{1-y^2}(x+7y^3)\,\mathrm dx\,\mathrm dy\]or, in the opposite order,
\[I=\int_{-1}^1\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}(x+7y^3)\,\mathrm dy\,\mathrm dx\]

Answer 6

Ah, that makes sense to me now, thank you both.

Answer 7

Quick question though.
Why does \(y=\sqrt{1-x^2}\) only equal half of a circle when you can do:
\[y^2=1-x^2 \rightarrow y^2+x^2=1\]
if you square both sides of the equation.
I know its a simple question but its kind of throwing me off.

Answer 8

That's because \(y^2=1-x^2\) as two solutions for \(y\), both \(y=\sqrt{1-x^2}\) and \(y=-\sqrt{1-x^2}\). This all comes down to the fact that \(f(x)=x^2\) is not one-to-one.