# Use Polar coordinates to evaluate the Double integral \iint _R (x^2+y^2+3) dA, where R is the...

## Question:

Use Polar coordinates to evaluate the Double integral {eq}\iint _R (x^2+y^2+3) dA, {/eq} where {eq}R {/eq} is the circle of radius {eq}2 {/eq} centered at the origin.

## Double integral:

If an integral is given in the form:

{eq}\displaystyle I = \iint_R f(x,y) \, dA {/eq}, where R is the region bounded between the given curves.

Then, the polar form of the given integral is:

{eq}\displaystyle I = \iint_D f(r,\theta)\cdot r \, dr\,d\theta {/eq}

Where:

{eq}x = r\cos\theta \\ y = r\sin\theta \\ r = \sqrt{x^2+y^2} {/eq}

These are the polar terms of the rectangular coordinates.

The given integral is:

{eq}\displaystyle I = \iint _R (x^2+y^2+3) \, dA {/eq}

Where, {eq}R: x^2 + y^2 = 4 {/eq}

We have:

{eq}f(x, y) = x^2+y^2+3 {/eq}

Using the polar coordinates, we get:

{eq}f(r, \theta) = r^2 + 3 {/eq}

And:

{eq}r^2 = 4 \\ r = \pm 2 {/eq}

Radius cannot be negative, so we have:

{eq}0 \leq r \leq 2 {/eq}

Where:

{eq}0 \leq \theta \leq 2\pi {/eq}

Evaluating the integral by using the polar coordinates:

{eq}\displaystyle I = \iint _R (x^2+y^2+3) \, dA \\ \displaystyle I = \int _{0}^{2\pi}\int _{0}^{2} (r^2+3)\cdot r \, dr\,d\theta \\ \displaystyle I = \int _{0}^{2\pi}\int _{0}^{2} r^3 + 3r \, dr\,d\theta \\ \displaystyle I = \int _{0}^{2\pi}\left[ \frac{r^4}{4} + \frac{3r^2}{2} \right ]_{0}^{2} \,d\theta \\ \displaystyle I = \int _{0}^{2\pi}\left[ \frac{2^4}{4} + \frac{3(2)^2}{2} \right ] \,d\theta \\ \displaystyle I = \int _{0}^{2\pi}\left[ 10 \right ] \,d\theta \\ \displaystyle I = 10\int _{0}^{2\pi} \,d\theta \\ \displaystyle I = 10\left[ \theta\right] _{0}^{2\pi} \\ \displaystyle I = 10\left[ 2\pi \right]\\ I = 20\pi {/eq}

Thus,

{eq}\displaystyle \boxed{\iint _R (x^2+y^2+3) \, dA = 20\pi} {/eq}