please help w this! Its a calculus question! look at the picture for the problem,

Since you mentioned calculus, perhaps you're supposed to find the area by integration.The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines [tex]x=\pm3\sqrt2[/tex] and [tex]y=\pm3\sqrt2[/tex].Let [tex]R[/tex] be the region bounded by the line [tex]x=3\sqrt2[/tex] and the circle [tex]x^2+y^2=36[/tex] (the rightmost blue region). The right side of the circle can be expressed in terms of [tex]x[/tex] as a function of [tex]y[/tex]:[tex]x^2+y^2=36\implies x=\sqrt{36-y^2}[/tex]Then the area of this circular segment is[tex]\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy[/tex][tex]=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy[/tex]Substitute [tex]y=6\sin t[/tex], so that [tex]\mathrm dy=6\cos t\,\mathrm dt[/tex][tex]=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt[/tex][tex]=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18[/tex]Then the area of the entire blue region is 4 times this, a total of [tex]\boxed{36\pi-72}[/tex].Alternatively, you can compute the area of [tex]R[/tex] in polar coordinates. The line [tex]x=3\sqrt2[/tex] becomes [tex]r=3\sqrt2\sec\theta[/tex], while the circle is given by [tex]r=6[/tex]. The two curves intersect at [tex]\theta=\pm\dfrac\pi4[/tex], so that[tex]\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta[/tex][tex]=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18[/tex]so again the total area would be [tex]36\pi-72[/tex].Or you can omit using calculus altogether and rely on some basic geometric facts. The region [tex]R[/tex] is a circular segment subtended by a central angle of [tex]\dfrac\pi2[/tex] radians. Then its area is[tex]\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18[/tex]so the total area is, once again, [tex]36\pi-72[/tex].An even simpler way is to subtract the area of the square from the area of the circle.[tex]\pi6^2-(6\sqrt2)^2=36\pi-72[/tex]