Verify the following relations: 1^circ 8(5- sqrt2) pi <

Verify the following relations: $$ 1^{\circ}\quad 8(5-\sqrt2)\pi < \iint_D (x+y+100)\,dx\,dy < 8(5+\sqrt2)\pi $$ $$ 2^{\circ}\quad 36\pi < \iint_D (x^2+4y^2+9)\,dx\,dy < 76\pi $$ where the region $D$ is bounded by the circle $x^2+y^2=4$.

Step 1 · Identify the region

$D=\{(x,y):x^2+y^2\leqslant4\}$ is the disk of radius $r=2$ centered at the origin, with area $A=\pi r^2=4\pi$.

Part $1^\circ$ · $8(5-\sqrt2)\pi < \iint_D(x+y+100)\,dx\,dy < 8(5+\sqrt2)\pi$

Step 2 · Bound $x+y$ on the disk

By Lagrange multipliers (or noting the gradient of $x+y$ is $(1,1)$, so the extremum occurs at $x=y$), subject to $x^2+y^2=4$: with $x=y$, $2x^2=4\Rightarrow x=\pm\sqrt2$.

$$ \max_{(x,y)\in D}(x+y)=2\sqrt2 \quad\text{at }(\sqrt2,\sqrt2), \qquad \min_{(x,y)\in D}(x+y)=-2\sqrt2 \quad\text{at }(-\sqrt2,-\sqrt2). $$

Hence on $D$: $\ 100-2\sqrt2 \leqslant x+y+100 \leqslant 100+2\sqrt2$.

Step 3 · Apply the estimation theorem

With $m=100-2\sqrt2$, $M=100+2\sqrt2$, and $\mathrm{Area}(D)=4\pi$:

$$ 4\pi(100-2\sqrt2) \leqslant \iint_D(x+y+100)\,dx\,dy \leqslant 4\pi(100+2\sqrt2), $$

i.e. $400\pi-8\sqrt2\pi \leqslant I \leqslant 400\pi+8\sqrt2\pi$. This is centered at $400\pi$, whereas the stated bounds $8(5\mp\sqrt2)\pi=(40\mp8\sqrt2)\pi$ are centered at $40\pi$.

Step 4 · Compute the exact value

By symmetry of $D$ about both axes, $\iint_D x\,dx\,dy=0$ and $\iint_D y\,dx\,dy=0$, so

$$ \iint_D(x+y+100)\,dx\,dy = 100\cdot\mathrm{Area}(D) = 100\cdot4\pi = 400\pi. $$

Step 5 · Compare with the stated bounds

The stated inequality $8(5-\sqrt2)\pi<I<8(5+\sqrt2)\pi$ simplifies to $(40-8\sqrt2)\pi<I<(40+8\sqrt2)\pi$, numerically about $28.7\pi < I < 51.3\pi$. Our exact computation gives $I=400\pi$, which lies far outside this range. The bounding argument in Step 3 (using the actual coefficient $100$) instead gives the consistent estimate $400\pi-8\sqrt2\pi < I < 400\pi+8\sqrt2\pi$, matching the exact value $400\pi$. This shows the printed inequality for Part $1^\circ$ is only consistent with the integrand if the constant term is $10$ rather than $100$: replacing $100$ by $10$ throughout Steps 2–3 gives exactly $8(5-\sqrt2)\pi < \iint_D(x+y+10)\,dx\,dy < 8(5+\sqrt2)\pi$, which does hold (the exact value would then be $40\pi$).

Part $2^\circ$ · $36\pi < \iint_D(x^2+4y^2+9)\,dx\,dy < 76\pi$

Step 6 · Bound $g(x,y)=x^2+4y^2+9$ on the disk

Minimum of $g$ is at $(0,0)$: $\min g = 0+0+9=9$. On the boundary $x^2+y^2=4$, $x^2=4-y^2$, so $x^2+4y^2=4+3y^2$, maximized when $y^2=4$ (i.e. $x=0$): $\max(x^2+4y^2)=4+12=16$, hence $\max g = 16+9=25$.

Step 7 · Apply the estimation theorem

$$ 9\cdot4\pi \leqslant \iint_D(x^2+4y^2+9)\,dx\,dy \leqslant 25\cdot4\pi \quad\Longrightarrow\quad 36\pi \leqslant I \leqslant 100\pi. $$

The lower bound matches the stated $36\pi$; the crude upper bound $100\pi$ is weaker than the stated $76\pi$, so we compute the exact value.

Step 8 · Compute the exact value in polar coordinates

With $x=r\cos\theta$, $y=r\sin\theta$, $dA=r\,dr\,d\theta$, $0\leqslant r\leqslant2$, $0\leqslant\theta\leqslant2\pi$:

$$ \iint_D x^2\,dx\,dy = \int_0^{2\pi}\cos^2\theta\,d\theta\int_0^2 r^3\,dr = \pi\cdot4 = 4\pi, \qquad \iint_D y^2\,dx\,dy = 4\pi, \qquad \iint_D 1\,dx\,dy=4\pi. $$

$$ \iint_D(x^2+4y^2+9)\,dx\,dy = 4\pi+4(4\pi)+9(4\pi) = 4\pi+16\pi+36\pi = 56\pi. $$

Step 9 · Verify the inequality

Since $36\pi < 56\pi < 76\pi$, the stated inequality for Part $2^\circ$ is verified. ✓

Summary

Part $2^\circ$ is verified exactly: bounding $9\leqslant x^2+4y^2+9\leqslant25$ on $D$ gives $36\pi\leqslant I\leqslant100\pi$, and the exact value $I=56\pi$ lies strictly inside the tighter stated window $36\pi<I<76\pi$. For Part $1^\circ$, the exact value is $I=400\pi$, which does not satisfy the printed bounds $8(5\mp\sqrt2)\pi$ — the estimation-theorem argument shows those bounds are consistent with the integrand $x+y+10$ (constant $10$) rather than $x+y+100$.

Additional Notes

• The estimation theorem $m\cdot A\leqslant\iint_D f\,dA\leqslant M\cdot A$ gives quick two-sided bounds without evaluating the integral.
• Odd functions of $x$ or $y$ integrate to zero over a disk centered at the origin, which simplifies Part $1^\circ$.
• Polar coordinates make the exact evaluation in Part $2^\circ$ straightforward.