Question

∫∫xy^3dxdy d: y^2=1-x, x ⩾0

Answer 1

Ответ: 0.

Пошаговое объяснение:

$\iint {xy^3dxdy}=\int\limits^1_0 dx \int\limits^{\sqrt{1-x} }_{-\sqrt{1-x} }{xy^3\, dy}=\int\limits^1_0 xdx (\frac{y^4}{4} )\bigg|\limits^{\sqrt{1-x} }_{-\sqrt{1-x}}=\int\limits^1_0 xdx (\frac{(1-x)^2-(1-x)^2}{4} )=\int\limits^1_0 0 \, dx=0$

Answer 2

To evaluate the double integral ∬xy^3 dxdy over the region D defined by y^2 = 1 - x and x ≥ 0, we first need to determine the limits of integration.

1. Finding the limits of integration: Since y^2 = 1 - x, we can rearrange it to get x = 1 - y^2. As x ≥ 0, we set 1 - y^2 ≥ 0, which implies y^2 ≤ 1. Taking the square root, we get -1 ≤ y ≤ 1.

Also, since x ≥ 0, we have 0 ≤ x ≤ 1.

So the limits of integration are: 0 ≤ x ≤ 1 -1 ≤ y ≤ 1

2. Evaluating the integral: Now, we can evaluate the double integral:

∬xy^3 dxdy = ∫[from -1 to 1] ∫[from 0 to 1] xy^3 dx dy

Let's perform the integration step by step:

∫[from 0 to 1] xy^3 dx = y^3 * ∫[from 0 to 1] x dx = y^3 * [x^2 / 2] | (from 0 to 1) = y^3 * (1^2 / 2) - y^3 * (0^2 / 2) = y^3 * (1/2) - 0 = y^3 / 2

Now, integrate with respect to y:

∫[from -1 to 1] y^3 / 2 dy = (1/2) * ∫[from -1 to 1] y^3 dy = (1/2) * [(y^4 / 4) | (from -1 to 1)] = (1/2) * [(1/4) - ((-1)^4 / 4)] = (1/2) * [(1/4) - (1/4)] = (1/2) * 0 = 0

So, the result of the double integral ∬xy^3 dxdy over the region D is 0.