Question

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

Answer 1

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
`x=\pm3\sqrt2` and
`y=\pm3\sqrt2`.

Let
`R` be the region bounded by the line
`x=3\sqrt2` and the circle
`x^2+y^2=36` (the rightmost blue region). The right side of the circle can be expressed in terms of
`x` as a function of
`y`:

`x^2+y^2=36\implies x=√(36-y^2)`

Then the area of this circular segment is

`\displaystyle\iint_R\mathrm dA=\int_(-3\sqrt2)^(3\sqrt2)\int_(3\sqrt2)^(√(36-y^2))\,\mathrm dx\,\mathrm dy`

`=\displaystyle\int_(-3\sqrt2)^(3\sqrt2)(√(36-y^2)-3\sqrt2)\,\mathrm dy`

Substitute
`y=6\sin t`, so that
`\mathrm dy=6\cos t\,\mathrm dt`

`=\displaystyle\int_(-\pi/4)^(\pi/4)6\cos t(√(36-(6\sin t)^2)-3\sqrt2)\,\mathrm dt`

`=\displaystyle\int_(-\pi/4)^(\pi/4)(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18`

Then the area of the entire blue region is 4 times this, a total of
`\boxed{36\pi-72}`.

Alternatively, you can compute the area of
`R` in polar coordinates. The line
`x=3\sqrt2` becomes
`r=3\sqrt2\sec\theta`, while the circle is given by
`r=6`. The two curves intersect at
`\theta=\pm\frac\pi4`, so that

`\displaystyle\iint_R\mathrm dA=\int_(-\pi/4)^(\pi/4)\int_(3\sqrt2\sec\theta)^6r\,\mathrm dr\,\mathrm d\theta`

`=\displaystyle\frac12\int_(-\pi/4)^(\pi/4)(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18`

so again the total area would be
`36\pi-72`.

Or you can omit using calculus altogether and rely on some basic geometric facts. The region
`R` is a circular segment subtended by a central angle of
`\frac\pi2` radians. Then its area is

`\frac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18`

so the total area is, once again,
`36\pi-72`.

An even simpler way is to subtract the area of the square from the area of the circle.

`\pi6^2-(6\sqrt2)^2=36\pi-72`