To solve the double integral over the region enclosed by the triangle with vertices at (0,0), (1,0), and (1,1), we first need to set up the limits of integration.

The limits of x are from 0 to 1 (from the x-coordinates of the points), and for each x, y goes from 0 to x (the line y=x is the hypotenuse of the triangle).

So, the double integral becomes:

∫ (from 0 to 1) ∫ (from 0 to x) (x^2 + y^2) dy dx.

First, we integrate with respect to y:

∫ (from 0 to 1) [x^2*y + (1/3)*y^3] (from 0 to x) dx = ∫ (from 0 to 1) (x^3 + (1/3)*x^3) dx = ∫ (from 0 to 1) (4/3)*x^3 dx.

Then, we integrate with respect to x:

[(4/12)*x^4] (from 0 to 1) = (1/3) - 0 = 1/3.

So, the value of the double integral over the given region is 1/3.

Question

To solve the double integral over the region enclosed by the triangle with vertices at (0,0), (1,0), and (1,1), we first need to set up the limits of integration.

The limits of x are from 0 to 1 (from the x-coordinates of the points), and for each x, y goes from 0 to x (the line y=x is the hypotenuse of the triangle).

So, the double integral becomes:

∫ (from 0 to 1) ∫ (from 0 to x) (x^2 + y^2) dy dx.

First, we integrate with respect to y:

∫ (from 0 to 1) [x^2*y + (1/3)*y^3] (from 0 to x) dx = ∫ (from 0 to 1) (x^3 + (1/3)*x^3) dx = ∫ (from 0 to 1) (4/3)*x^3 dx.

Then, we integrate with respect to x:

[(4/12)*x^4] (from 0 to 1) = (1/3) - 0 = 1/3.

So, the value of the double integral over the given region is 1/3.

Knowee AI · Accepted Answer

To solve the double integral over the region enclosed by the triangle with vertices at (0,0), (1,0), and (1,1), we first need to set up the limits of integration.

The limits of x are from 0 to 1 (from the x-coordinates of the points), and for each x, y goes from 0 to x (the line y=x is the hypotenuse of the triangle).

So, the double integral becomes:

∫ (from 0 to 1) ∫ (from 0 to x) (x^2 + y^2) dy dx.

First, we integrate with respect to y:

∫ (from 0 to 1) [x^2*y + (1/3)*y^3] (from 0 to x) dx = ∫ (from 0 to 1) (x^3 + (1/3)*x^3) dx = ∫ (from 0 to 1) (4/3)*x^3 dx.

Then, we integrate with respect to x:

[(4/12)*x^4] (from 0 to 1) = (1/3) - 0 = 1/3.

So, the value of the double integral over the given region is 1/3.

Evaluate ∫∫(xଶ + yଶ)dxdy over the region enclosed by thetriangle having vertices at (0, 0), (1,0), (1,1).

Question

Evaluate $\int\int (x^2 + y^2) \, dx \, dy$ over the region enclosed by the triangle having vertices at $(0, 0)$ , $(1, 0)\,$ and $(1, 1)$ .

Solution

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