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Let \(R\) be the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1,\) where \(a>0\) and \(b>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v\) Find the area of \(R\)

Short Answer

Expert verified
Answer: The area of the region bounded by the ellipse is \(A = ab\pi\).

Step by step solution

01

Apply the Transformation

Replace \(x\) with \(au\) and \(y\) with \(bv\) in the equation of the ellipse: \((au)^2/a^2 + (bv)^2/b^2 = 1\) which simplifies to \(u^2 + v^2 = 1\). This is the equation of a unit circle with center at \((0,0)\) and radius \(1\).
02

Calculate the Area of the Unit Circle

The area of a circle is given by the formula \(A = \pi r^2\). Since the radius of our transformed circle is \(1\), the area is \(A = \pi(1^2) = \pi\).
03

Undo the Transformation

In step 1, we applied the transformation \(x = au\) and \(y = bv\). To find the area of the original ellipse, we need to undo this transformation. We do this by multiplying the area of the unit circle by the inverse of the transformation, which is \(ab\). Therefore, the area of the ellipse is \(A= ab \cdot \pi\).
04

Final Answer

The area of the region bounded by the ellipse \(x^2/a^2 + y^2/b^2 = 1\) is \(A = ab\pi\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ellipses and circles
Ellipses and circles are fundamental shapes in geometry and calculus. A circle is essentially a special type of ellipse where both of its axes are equal, meaning the radius is the same in every direction. In the equation of a circle, \[x^2 + y^2 = r^2\]"\[r\]" represents the radius of the circle. For an ellipse, the formula becomes more general:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]where "\[a\]" and "\[b\]" are the semi-major and semi-minor axes, respectively. This equation describes an ellipse centered at the origin. If \[a eq b\], the shape is stretched along one axis. Understanding this relationship is pivotal in converting ellipses to circles through appropriate transformations.
Coordinate transformation
Coordinate transformation is a powerful mathematical technique used to simplify complex equations or geometric shapes by changing the coordinate system. In this exercise, we utilized the transformation formulas:
  • \[x = au\]
  • \[y = bv\]
These transformations allow us to map an ellipse described by \(x^2/a^2 + y^2/b^2 = 1\) to a unit circle given by \(u^2 + v^2 = 1\). Essentially, we are scaling the "x" and "y" coordinates by "a" and "b", turning the more complicated ellipse into a simple circle. This conversion makes many calculations, such as area finding, easier to work with. Once we understand the new, transformed space, we can calculate easily and then convert our results back into the original system.
Area calculation
Calculating areas of geometric figures is a key concept in calculus. For areas within transformed coordinates, like in this problem, the steps are often:
  • Transform the figure (ellipse) into a simpler shape (unit circle).
  • Calculate the area of the simple shape using standard formulae.
  • Undo the transformation to achieve the correct area for the original shape.
For a unit circle with radius 1, the area is straightforward:\[A = \pi r^2 = \pi \]To find the original area's size when mapped back to the ellipse, we adjust by the inverse of our transformation, multiplying by "\(ab\)" (since both "a" and "b" scale the dimensions). This leads to:\[A_{ellipse} = ab \cdot \pi\]Mastering these steps aids in understanding more complex area problems.

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Most popular questions from this chapter

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of a right circular cone with height \(h\) and base radius \(r\)

Let \(D\) be the solid bounded by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1,\) where \(a>0, b>0,\) and \(c>0\) are real numbers. Let \(T\) be the transformation \(x=\)au, \(y=b v, z=c w\) Evaluate \(\iiint_{D}|x y z| d A\)

A thin (one-dimensional) wire of constant density is bent into the shape of a semicircle of radius \(a\). Find the location of its center of mass. (Hint: Treat the wire as a thin halfannulus with width \(\Delta a,\) and then let \(\Delta a \rightarrow 0\).)

A cake is shaped like a hemisphere of radius 4 with its base on the \(x y\) -plane. A wedge of the cake is removed by making two slices from the center of the cake outward, perpendicular to the \(x y\) -plane and separated by an angle of \(\varphi\) a. Use a double integral to find the volume of the slice for \(\varphi=\pi / 4 .\) Use geometry to check your answer. b. Now suppose the cake is sliced by a plane perpendicular to the \(x y\) -plane at \(x=a > 0 .\) Let \(D\) be the smaller of the two pieces produced. For what value of \(a\) is the volume of \(D\) equal to the volume in part (a)?

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the transformation \(T: x=g(u, v), y=h(u, v)\) is linear in \(u\) and \(v,\) then the Jacobian is a constant. b. The transformation \(x=a u+b v, y=c u+d v\) generally maps triangular regions to triangular regions. c. The transformation \(x=2 v, y=-2 u\) maps circles to circles.

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