Sketch the region R of a lamina which is bounded by the curve y=sqrt*(1-x^2) in the first quadrant.

y=sqrt*(1-x^2) figure

Answer 1

[Image Analyst]

What is a lamina? What defines "the first quadrant"? What range of x do you want to make y for? You can't sketch anything meaningful if x goes from -infinity to +infinity. What range of x do you want to use? Maybe -2 to 2 for example

x = linspace(-2, 2, 500);
y=sqrt(1-x.^2) ;
plot(x, y, 'b-', 'LineWidth', 2);
xlabel('x', 'FontSize', 24);
ylabel('y', 'FontSize', 24);
grid on;

If you wanted x to be only in the first quadrant, what would you have to change in the code?

[sensei]

if function of the lamina is given by delta(x,y)=cos(x.^2+y.62). how to set up the double integral to find Cartesian coordinates. can you help me? this has related to this question.

[Image Analyst]

You didn't answer my question yet. What is a lamina? To me it means a layer but I don't know how it applies in your context of plotting a math equation. I also don't know how a double integral would be involved in your latest equation. You could make a 2D array called delta using meshgrid and then use imshow() to display delta as an image. Do you want that? That's not using a double integral though.

x = linspace(-pi, pi, 500);
y = linspace(-pi, pi, 500);
[X, Y] = meshgrid(x, y);
delta = cos(X.^2+Y.^2);
imshow(delta, []);