Evaluating Integral Geometric Interpretation And Coordinate Transformation

Hey guys! Today, we're diving into a super interesting integral problem that involves visualizing regions, understanding geometric interpretations, and transforming coordinate systems. Buckle up, because this is going to be a fun ride!

The Integral Challenge

We're faced with the following double integral:

$$\int_{-1}^1 \int_0^{\sqrt{1-x^2}} 1-y^2 dy dx$$

Our mission, should we choose to accept it (and we totally do!), is to:

(a) Sketch the region of integration. (b) Give a geometric interpretation of the integral using a 3-dimensional sketch. (c) Transform the integral to a more suitable coordinate system for easier evaluation.

Let's break it down step by step. We will use a casual and friendly tone, like saying "guys" or other slang, so it feels natural and conversational. We will focus on creating high-quality content and providing value to the readers.

(a) Sketching the Region of Integration

Okay, first things first, let's figure out what region we're actually integrating over. To sketch the region of integration, we need to analyze the limits of integration. Remember, the inner integral (with respect to y in this case) defines the vertical boundaries, and the outer integral (with respect to x) defines the horizontal boundaries.

• The Outer Limits: The x limits go from -1 to 1. This tells us our region is bounded on the left by the vertical line x = -1 and on the right by the vertical line x = 1.

• The Inner Limits: The y limits are a bit more interesting. y goes from 0 to $\sqrt{1-x^2}$. Let's analyze that upper limit:

  $$y = \sqrt{1-x^2}$$

  Squaring both sides, we get:

  $$y^2 = 1 - x^2$$

  Rearranging, we have:

  $$x^2 + y^2 = 1$$

  Ah-ha! This is the equation of a circle with a radius of 1, centered at the origin (0, 0). But hold on! We need to consider the original limit y = $\sqrt{1-x^2}$. Since the square root function always returns a non-negative value, y is greater than or equal to 0. This means we only have the upper half of the circle.

  So, combining this information, our y limits describe the region bounded below by the x-axis (y = 0) and above by the upper half of the unit circle (x² + y² = 1).

• Putting it Together: We've got a region bounded by x = -1, x = 1, y = 0, and the upper half of the unit circle. If you visualize this, you'll see it's a semi-circle in the xy-plane, lying above the x-axis.

To nail this down, think of it this way: Sketching the region of integration involves understanding how the limits define the boundaries. The x limits are straightforward vertical lines, and the y limits describe the upper half of a circle. Combining these, we get our semi-circular region. Make sure your sketch the region of integration accurately shows this semi-circle; it’s the foundation for understanding the rest of the problem. Understanding sketch the region of integration here is the key to unlocking the rest of the problem, so make sure you’ve got a good visual in your head!

(b) Geometric Interpretation in 3D

Now for the fun part – giving a geometric interpretation of the integral! This is where we get to visualize what the integral actually means in 3D space. Remember, a double integral can be interpreted as the volume under a surface.

• The Surface: Our integrand is 1 - y². This represents a surface in 3D space. Notice that it only depends on y. This means that for any fixed value of y, the value of the function 1 - y² is constant along the x-direction. If we fix y, the surface looks like a curve in the yz-plane extruded along the x-axis. Specifically, the cross-sections parallel to the yz-plane are parabolas opening downwards.

• Visualizing the Surface: Imagine the parabola z = 1 - y² in the yz-plane. It intersects the z-axis at z = 1 (when y = 0) and intersects the y-axis at y = -1 and y = 1 (when z = 0). Now, imagine stretching this parabola along the x-axis. You'll get a curved surface that looks like a curved tunnel or a Quonset hut shape.

• The Volume: The integral $\int_{-1}^1 \int_0^{\sqrt{1-x^2}} 1-y^2 dy dx$ calculates the volume under this curved surface (z = 1 - y²) and above the semi-circular region we sketched in part (a). Think of it as slicing the 3D space with our semi-circular region and then finding the volume of the solid formed between the xy-plane and the surface.

  To really nail this geometric intuition, try to visualize how the dy dx part of the integral corresponds to infinitesimal areas in the xy-plane. For each small area element, we’re multiplying it by the height of the surface (1 - y²) to get a tiny volume. The integral then sums up all these tiny volumes to give us the total volume.

  When thinking about geometric interpretation, always focus on the function being integrated and the region over which you're integrating. In this case, the function 1 - y² forms a curved surface, and the integral is essentially summing up the volume under that surface over our semi-circular region. This geometric interpretation transforms the abstract calculation into a concrete visual in 3D space, providing a deeper understanding of what the integral represents. Grasping the geometric interpretation isn't just about sketching a picture; it's about connecting the mathematical expression to a tangible volume in space.

(c) Transforming to Polar Coordinates

Alright, now let's talk about making our lives easier. Evaluating the integral in rectangular coordinates (i.e., with x and y) might be a bit messy. But guess what? Our region of integration is a semi-circle – a perfect candidate for polar coordinates! Transforming to polar coordinates will simplify both the region and potentially the integrand, making the integration process much smoother.

• Recall Polar Coordinates: Remember, polar coordinates use the radius r and the angle θ to describe points in the plane:

  • x = rcos(θ)
  • y = rsin(θ)
  • x² + y² = r²
  • dA = dx dy = r dr dθ (the Jacobian transformation)

  The last one is crucial – when we change variables, we need to account for the change in area element. dA becomes r dr dθ in polar coordinates.

• Transforming the Region: In polar coordinates, our semi-circular region is beautifully simple:

  • Since we have the upper half of the unit circle, the radius r goes from 0 to 1.
  • The angle θ goes from 0 to π (covering the upper half of the circle).

• Transforming the Integrand: Now, let's transform the 1 - y² part of the integral:

  • Substitute y = rsin(θ): 1 - y² = 1 - (rsin(θ))² = 1 - r²sin²(θ)*

• The Transformed Integral: Putting it all together, our integral becomes:

  $$\int_{0}^{\pi} \int_{0}^{1} (1 - r^2sin^2(\theta)) r dr d\theta$$

  Notice how the limits of integration are now constants! This is a major win. The integrand is a bit more complex, but we've eliminated the square root in the limits, which is a huge step forward.

  Transforming to polar coordinates is a technique that leverages symmetry to simplify integrals. The key here is to recognize that our semi-circular region translates beautifully into constant limits for r and θ. This simplification alone often makes the integral much easier to handle. While the integrand 1 - r²sin²(θ) looks a bit more complicated than 1 - y², it’s often manageable, and the constant limits of integration are a significant advantage. Master the art of transforming to polar coordinates and you’ll have a powerful tool in your calculus arsenal.

  To make sure you've got it, think about why polar coordinates worked so well here. The circular symmetry of the region made the transformation incredibly effective. If we had a different region (like a square), polar coordinates might not be the best choice. Part of becoming a calculus ninja is recognizing when transforming to polar coordinates is the right move.

Conclusion

So, there you have it! We've successfully tackled this integral by sketching the region of integration, visualizing it geometrically in 3D, and transforming to polar coordinates to make the evaluation easier. This problem perfectly illustrates how different mathematical concepts – geometry, calculus, and coordinate transformations – come together to solve a single problem.

Remember, guys, practice makes perfect! Keep exploring integrals, keep visualizing, and keep those problem-solving skills sharp. You've got this!