<s>[INST] Solve the area on quadrant one given by the two curves $x^2+y^2=9$ and $ 4x^2+3y^2=36$
I'm having difficulties solving this, but I was told the answer is $\frac{3}{4}\pi$
However, when I equate the two and get $x = 3$ I try to solve them but can't get the correct answer after integration. I was thinking if these are like $2$ circle problem and then apply definite integrals but then again I can't get the correct answer. 
 [/INST] The required region (shaded portion) is given below:

The upper curve is given by $y=\sqrt{\frac{36-4x^2}{3}}$ and the lower curve is $y=\sqrt{9-x^2}$. Thus, the area is given by
$$A=\int_{0}^3\sqrt{\frac{36-4x^2}{3}}dx-\frac{\pi 3^2}{4}$$
Using Wolfram Alpha, 
$$\int_{0}^3\sqrt{\frac{36-4x^2}{3}}dx=\frac{3\sqrt{3}\pi}{2}$$ but as you have noted the integral is $\frac{\text{area of the ellipse}}{4}=\frac{\pi(3)\sqrt{12}}{4}=\frac{3\sqrt{3}\pi}{2}$. Hence, the required area is
$$A=\frac{3\sqrt{3}\pi}{2}-\frac{9\pi}{4}=\frac{(6\sqrt{3}-9)\pi}{4}$$
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