Double integral?

First, you cannot separate the integrals, since y has bounds that are dependent on x:

∫ [a to b] ∫ [c to d] f(x,y) dy dx

can be split into 2 separate integrals when bounds a, b, c, d are constant

and f(x,y) = g(x) * h(y)

∫ [a to b] ∫ [c to d] f(x,y) dy dx = ∫ [a to b] g(x) dx * ∫ [c to d] h(y) dy

Since this is not the case here, double integral should be written:

∫ [0 to 2] ∫ [0 to x] f(√(x²+y²)) dy dx

(do not put dx in front of the second integral)

Second, the function being integrated is NOT √(x²+y²). It is f(√(x²+y²)), which is not the same, unless f is a single variable function (let's use t) and is: f(t) = t, then f(√(x²+y²)) IS = √(x²+y²). So if function f is not defined anywhere, do not write f(√(x²+y²)). If you want to integrate √(x²+y²), then use that, not f(√(x²+y²)).

But perhaps function is f(x,y) = √(x²+y²), and integral is ∫ [0 to 2] ∫ [0 to x] f(x,y) dy dx

In that case, this is equivalent to ∫ [0 to 2] ∫ [0 to x] √(x²+y²) dy dx

but it is not equivalent to ∫ [0 to 2] ∫ [0 to x] f(√(x²+y²)) dy dx

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Now to sketch the region:

Graph y = 0 and y = x. Shade region in between these lines, for x on the interval [0, 2]

https://imgur.com/29MOZ7Z

The function f(x,y) = √(x²+y²) is a cone that lies above the xy-plane.

So the integral ∫ [0 to 2] ∫ [0 to x] √(x²+y²) dy dx is used to calculate the volume between that cone and xy-plane, above the triangular region that we sketched above.

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∫ [0 to 2] ∫ [0 to x] √(x²+y²) dy dx

= ∫ [0 to 2] { 1/2 ( y √(x²+y²) + x² ln(y + √(x²+y²)) ) } | [0 to x] dx

= ∫ [0 to 2] 1/2 [ ( x √(2x²) + x² ln(x + √(2x²)) ) − ( 0 + x² ln(0 + √(x²)) ) ] dx

= ∫ [0 to 2] 1/2 (√2 + ln(1+√2)) x²

= 1/6 (√2 + ln(1+√2)) x³ [0 to 2]

= 4/3 (√2 + ln(1+√2))

≈ 3.060782865856850765379064065586

The region of integration is a right triangle whose vertices are

(0,0), (2,2), and (2,0). Very easy to draw.

The function sqrt(x^2 + y^2) should be thought of as a "z" standing up above the x-y plane. Its value is equal to the (horizontal) distance from the origin. In your problem, that surface is like the "under" side of a cone. Think of a downward-pointing cone whose tip is at (0,0,0). You are seeking the volume UNDERNEATH it but above the given triangle in the x-y plane. To sketch the figure whose volume you are seeking, note that its height at (0,0) is nothing, its height at (0,2) is 2, and its height at (2,2) is about 2.83.