Use an iterated integral to find the area of the region bounded by the graphs of the equations.

Use an iterated integral to find the area of the region...

Key Concepts

Coordinate Transformation and Jacobian

When direct integration in Cartesian coordinates becomes cumbersome, a coordinate transformation can simplify the process by converting the region into one with easier-to-describe limits. The Jacobian of the transformation accounts for the scaling of area elements, ensuring the integral remains equivalent to the original formulation.

Determining the Region of Integration

Setting up the region of integration requires understanding how the boundaries of the region, defined by given equations, constrain the variables. By solving one variable in terms of another, one can properly define the limits of integration for the iterated integral.

Area Computation via Double Integration

Calculating the area of a region can be achieved by setting up a double integral over the region, where the integration limits are determined by the boundary curves. For any region in the plane, this technique sums up 'infinitesimal' areas to yield the total area.

Iterated Integral

An iterated integral involves evaluating a multiple integral as a sequence of single-variable integrals, one after the other. This method is particularly useful for finding areas of regions by integrating one variable at a time according to limits defined by the region's boundary.

Transcript

00:01 For this problem, we are asked to use an iterated integral to find the area of the region bounded by the equation, x squared over a squared plus y squared over b squared equals 1.

00:10 So, to begin, we'll need to figure out what our boundaries for x and y should even be here.

00:16 Let's start out by setting y equals to 0, which would then mean that we'd need to have x squared over a squared equals 1, which means that we'd need to have x squared equals a squared, which then means that we'd need to have x equals plus or minus a.

00:33 So we can set up our integral as being, excuse me, not zero less than or equal to x, but rather we can set this up as x is between negative a and positive a.

00:46 Then in turn, if we have a non -zero x, we can rearrange our equation to try to get y in terms of x.

00:54 Particularly, we'd have that y squared over b squared, should be equal to 1 minus x squared over a squared, which then means that y squared would be equal to b squared minus b squared over a squared times x squared, which would mean that y is going to equal plus or minus the square root of b squared minus b squared over a squared times x squared.

01:21 So we can then define y in terms of x by saying that y should be between negative, and i'll write this just sort of factoring out that b squared, making my life a little bit easier here.

01:32 Negative b times the square root of 1 minus x squared over a squared, less than or equal to y, less than or equal to positive b times the square root of 1 minus x squared over a squared.

01:46 Which then means that we can set up our area here as being the integral from negative a up to a of the integral from negative, i'll just copy and paste these.

01:59 Let's see here.

02:00 Negative b times root 1 minus x squared over a squared up to be positive of that so i'll just copy and paste that up there and erase the negative sign and we're integrating over y first then over x so doing that y integral first this becomes just the integral from negative a to a of b root whoops i don't know why it pasted it all the way down there so it would be b root 1 minus x squared over a squared minus negative.

02:33 So then it would be plus b or plus b root 1 minus x squared over a squared.

02:41 So we can write this as being 2 times b root 1 minus x squared over a squared dx or we can bring that 2b out front.

02:49 So it's 2b times the integral from negative a up to a of root 1 minus x squared over a squared.

02:59 So one moment here...

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