Answer to Question #140337 in Calculus for Promise Omiponle

Question

Evaluate the following integral after converting to polar coordinates: ∫(0 to 1)∫(y to sqrt(2-y^2)) (x+y) dx dy

Answer 1

$$Solution$$

STEP 1

Convert the upper bound of the integral for $x$ into polar coordinates. The result is a circle of radius $\sqrt 2$ centered at the origin.

$$x=\sqrt{2-y^2} \implies x^2=2-y^2 \implies x^2+y^2=2\\ r^2=2 \implies r=\sqrt 2$$

STEP 2

Convert the other bounds into polar coordinates. These tell us that the angle ranges from to $\frac{\pi}{4}$

I.e the region is a wedge of the circle of radius $\sqrt 2$, with an angle sweeping from zero to $\frac{\pi}{4}$

$$x=y \implies \theta=\frac{\pi}{4}\\ y=0 \implies \theta=0$$

STEP 3

Convert the function into polar coordinates

$$=x+y = r \cdot cos \theta + r \cdot sin \theta\\ =\intop_0^{\frac{\pi}{4}} \intop_0^{\sqrt{2}}( r \cdot cos \theta + r \cdot sin \theta)r\ \delta r\ \delta \theta\\ =\intop_0^{\frac{\pi}{4}} \intop_0^{\sqrt{2}}r^2( cos \theta + sin \theta)r\ \delta r\ \delta \theta\\$$

STEP 4

Integrate the function with respect to $r$

$$= \intop_0 ^{\frac{\pi}{4}}[\frac{1}{3}r^3(cos \theta+sin \theta)]_0^{\sqrt{2}} \delta \theta\\ = \intop_0 ^{\frac{\pi}{4}}\frac{2 \sqrt{2}}{3}(cos \theta+sin \theta) \delta \theta$$

STEP 5

Integrate the function with respect to theta

$$= [\frac{2 \sqrt{2}}{3}(sin \theta-cos \theta) ]_0^{\frac{\pi}{4}}=\frac{2 \sqrt{2}}{3}---->Answer$$