Double integral with polar coordinates?

Question

integral from o to pi/2, integral from 1 to 2 of sqrt(4-r^2*cos^2(theta)-r^2*sin^2(theta))rdrdtheta

if you could please help me by showing steps, that would be great! thanks!

Update:

sqrt(4 - r^2*cos^2(theta) - r^2*sin^2(theta)) r dr dtheta

thanks for letting me know!

Answer 1

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integral from 0 to pi/2, integral from 1 to 2 of sqrt(4-r^2*cos^2(theta)-r^2*sin^2(theta)…

Be aware that Y!A uses a text editor that does some destructive things, one of which is that, for any string of more than 30 characters you type, without a space, it will replace all the 'excess' characters with the ellipsis, "..." So you need to edit that expression to insert a space here and there so that there are no 30 non-space characters in a row anywhere in it.

Meanwhile, while you're repairing that, I can already see a simplifying step, if the chopped-off part doesn't have more stuff multiplying that last piece:

4 - r²cos²θ - r²sin²θ = 4 - r²(cos²θ + sin²θ) = 4 - r²

I'll wait to see what the whole thing is.

AddDet:

OK, so it's

: ½π: 2 : : : : : : : : : : : : : : : : : : : : ½π: 2

I = ∫ : ∫(4 - r²cos²θ - r²sin²θ) r dr dθ = ∫ : ∫(4 - r²) r dr dθ

: : 0: 1 : : : : : : : : : : : : : : : : : : : : : 0: 1

and the integral is separable, because the first integration (over r) carries no θ-dependency into the θ-integration. Then letting

u = r², du = 2r dr,

r={1,2} ⇒ u={1,4}

I = ∫∫(4 - r²) r dr dθ = [∫(4 - r²) r dr] ∫dθ = ½π ∫(4 - u) ½du =

: : : : : : : : : :4

¼π(4u - ½u²) | = ¼π([16-8] - [4-½]) = 9π/8

: : : : : : : : : :1