# 1. Compute (x^2 + y^2) dA over an annulus 1 x^2 + y^2 4. 2. Compute 0^1 ( 0^ ...

## Question:

1. Compute{eq}\displaystyle \ \int \int \cos (x^2 + y^2) \ dA {/eq} over an annulus {eq}\displaystyle \ 1 \leq x^2 + y^2 \leq 4. {/eq}

2. Compute{eq}\displaystyle \ \int_0^1 \left ( \int_0^{ \sqrt {1 - x^2} } \sin \left ( \sqrt {x^2 + y^2} \right ) \ dy \right ) \ dx {/eq}

3. Find the volume of the solid bounded below by the xy-plane and above by{eq}\displaystyle \ z = 2x; {/eq} on the sides by the cylinder{eq}\displaystyle \ (x - 1)^2 + y^2 = 1. {/eq}

## Double Integrals in Polar Coordinates:

If we have a double integral {eq}\displaystyle\iint_R f(x, y) \: dA, {/eq} where the region {eq}R {/eq} is a circle, annulus, or portion of a circle, and the integrand {eq}f(x, y) {/eq} is a function of {eq}x^2 + y^2, {/eq} then quite often it is easier to first convert the problem to polar coordinates, then evaluate. An annulus centered at the origin can be described as {eq}0 \leq \theta \leq 2\pi, \: a \leq r \leq b, {/eq} and a circle or portion of a circle is similarly described. Next use the fact that {eq}r^2 = x^2 + y^2 {/eq} to convert the integrand, and replace {eq}dA {/eq} with {eq}r \: dr \: d\theta. {/eq} The new integral is ready to go!

## Answer and Explanation:

1. Compute{eq}\displaystyle \ \iint \cos (x^2 + y^2) \ dA {/eq} over an

The annulus {eq}\displaystyle \ 1 \leq x^2 + y^2 \leq 4 {/eq} can be...

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from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14