Question

Evaluate the given integral by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 100x2 + 49y2 = 1

Answer 1

The area bounded by the sin (100x² + 49y²) dA is calculated by using integration is 2π (1 - cos 1).

An area bounded by the curve: When the two curves intersect then they bound the region is known as the area bounded by the curve.

Evaluate the integral by making an appropriate change of variables.

The equation of the ellipse is 100x² + 49y² = 1

Let cos t = 10x and sin t = 7y. Then we have

or x = 1/10 cost , y = 1/7 sint.

Then

=> 100 (1/10cost)^2 + 49 (1/7 sint)^2 = 1

=> cos^2 t + sin^2t = 1 which suggests a change of variable will be:

$\left \{ {{x(r,t) = r/10cost} \atop {y(r,t)=r/7sint}} \right.$

where 0≤r≤1 and 0≤t≤2$\pi$. Then we also have,

100x² + 49y² = r²

So,

=> ∫∫ sin (100x^2 + 49y^2)dA

     R

=> 2 ∫2$\pi$ ∫$1$  sinr^2 dr dt

       0     0

=> 4$\pi$ ∫$1$  rsinr^2 dr

          0    

                       

Now r² is replaced by r, then we get

=> 2$\pi$ ∫$2$  sinr^2 d(r^2)

          0

=> - 2$\pi$ cos r^2 $\left \{ {{r^2=1} \atop {r^2=0}} \right.$    

=> -2$\pi$ (cos1-cos0)

=> -2$\pi$ (-1 + cos1)

=> 2$\pi$(1-cos1)

Evaluating the integral by making an appropriate change of variables 2 sin(100x^2 + 49y^2) dA.

To learn more about the area bounded by the curve visit: brainly.com/question/24563834

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