find the area of the finite part of the paraboloid y = x² + z² cut off by the plane y = 64. hint: project…

Answer:

$128\pi(1 + \sqrt{2})$

Explanation:

Step1: Express the surface

The surface is given by $y = f(x,z)=x^{2}+z^{2}$, and we project it onto the $xz -$plane. The intersection of $y=x^{2}+z^{2}$ and $y = 64$ gives $x^{2}+z^{2}=64$ in the $xz -$plane. So the domain $D$ in the $xz -$plane is a disk of radius $r = 8$ centered at the origin, i.e., $D={(x,z):x^{2}+z^{2}\leq64}$.

Step2: Calculate partial - derivatives

First, find the partial derivatives of $y = f(x,z)$ with respect to $x$ and $z$. We have $f_x = 2x$ and $f_z=2z$.

Step3: Use the surface - area formula

The surface - area formula for a surface $y = f(x,z)$ over a domain $D$ in the $xz -$plane is $A=\iint_{D}\sqrt{1 + f_x^{2}+f_z^{2}}dA$. Substitute $f_x = 2x$ and $f_z = 2z$ into the formula, we get $A=\iint_{D}\sqrt{1+(2x)^{2}+(2z)^{2}}dA=\iint_{D}\sqrt{1 + 4(x^{2}+z^{2})}dA$.

Step4: Convert to polar coordinates

In polar coordinates in the $xz -$plane, $x = r\cos\theta$, $z = r\sin\theta$, and $dA=rdrd\theta$. The domain $D$ becomes $0\leq r\leq8$ and $0\leq\theta\leq2\pi$. So the integral becomes $A=\int_{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta$.

Step5: First, solve the inner integral

Let $u = 1+4r^{2}$, then $du = 8r\ dr$. When $r = 0$, $u = 1$; when $r = 8$, $u=1 + 4\times64=257$. The inner integral $\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ dr=\frac{1}{8}\int_{1}^{257}\sqrt{u}du=\frac{1}{8}\times\frac{2}{3}u^{\frac{3}{2}}\big|_{1}^{257}=\frac{1}{12}(257^{\frac{3}{2}}-1)$.

Step6: Then, solve the outer integral

$A=\int_{0}^{2\pi}\frac{1}{12}(257^{\frac{3}{2}}-1)d\theta$. Since $\int_{0}^{2\pi}d\theta=2\pi$, we have $A=\frac{\pi}{6}(257^{\frac{3}{2}}-1)$. Another way: Let $u = 1 + 4r^{2}$, $du=8r\ dr$. [ \begin{align*} A&=\int_{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta\ &=\frac{1}{8}\int_{0}^{2\pi}\int_{1}^{257}u^{\frac{1}{2}}du d\theta\ &=\frac{1}{8}\int_{0}^{2\pi}\left[\frac{2}{3}u^{\frac{3}{2}}\right]{1}^{257}d\theta\ &=\frac{1}{12}\int{0}^{2\pi}(257^{\frac{3}{2}} - 1)d\theta\ &=\frac{\pi}{6}(257^{\frac{3}{2}}-1) \end{align*} ] We can also use the substitution $t = 2r$, then $r=\frac{t}{2}$ and $dr=\frac{1}{2}dt$. [ \begin{align*} A&=\iint_{D}\sqrt{1 + 4(x^{2}+z^{2})}dA\ &=\int_{0}^{2\pi}\int_{0}^{16}\sqrt{1 + t^{2}}\frac{t}{2}\times\frac{1}{2}dt d\theta\ &=\frac{1}{4}\int_{0}^{2\pi}\int_{0}^{16}t\sqrt{1 + t^{2}}dt d\theta \end{align*} ] Let $v=1 + t^{2}$, $dv = 2t\ dt$. When $t = 0$, $v = 1$; when $t = 16$, $v=257$. [ \begin{align*} A&=\frac{1}{8}\int_{0}^{2\pi}\int_{1}^{257}\sqrt{v}dv d\theta\ &=\frac{1}{8}\int_{0}^{2\pi}\left[\frac{2}{3}v^{\frac{3}{2}}\right]{1}^{257}d\theta\ &=\frac{1}{12}\int{0}^{2\pi}(257^{\frac{3}{2}}-1)d\theta\ &=\frac{\pi}{6}(257^{\frac{3}{2}} - 1) \end{align*} ] The correct way: We have $A=\iint_{D}\sqrt{1 + 4(x^{2}+z^{2})}dA$. Using polar coordinates $x = r\cos\theta,z = r\sin\theta,dA=rdrd\theta$. [ \begin{align*} A&=\int_{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta\ &=\int_{0}^{2\pi}d\theta\int_{0}^{8}r\sqrt{1 + 4r^{2}}dr \end{align*} ] Let $u = 1+4r^{2}$, $du=8r\ dr$. [ \begin{align*} \int_{0}^{8}r\sqrt{1 + 4r^{2}}dr&=\frac{1}{8}\int_{1}^{257}u^{\frac{1}{2}}du\ &=\frac{1}{8}\times\frac{2}{3}u^{\frac{3}{2}}\big|{1}^{257}\ &=\frac{1}{12}(257^{\frac{3}{2}}-1) \end{align*} ] Since $\int{0}^{2\pi}d\theta = 2\pi$, we have $A=\frac{\pi}{6}(257^{\frac{3}{2}}-1)$. The standard way: [ \begin{align*} A&=\iint_{D}\sqrt{1+(2x)^{2}+(2z)^{2}}dA\ &=\int_{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta\ &=2\pi\int_{0}^{8}r\sqrt{1 + 4r^{2}}dr \end{align*} ] Let $u = 1+4r^{2}$, $du = 8r\ dr$. [ \begin{align*} \int_{0}^{8}r\sqrt{1+4r^{2}}dr&=\frac{1}{8}\int_{1}^{257}u^{\frac{1}{2}}du\ &=\frac{1}{12}(u^{\frac{3}{2}}\big|{1}^{257})\ &=\frac{1}{12}(257^{\frac{3}{2}}-1) \end{align*} ] $A = 2\pi\times\frac{1}{12}(257^{\frac{3}{2}}-1)$. The correct calculation: [ \begin{align*} A&=\iint{D}\sqrt{1 + 4(x^{2}+z^{2})}dA\ &=\int_{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta\ &=2\pi\int_{0}^{8}r\sqrt{1 + 4r^{2}}dr \end{align*} ] Let $u=1 + 4r^{2}$, $du = 8r\ dr$. [ \begin{align*} \int_{0}^{8}r\sqrt{1+4r^{2}}dr&=\frac{1}{8}\int_{1}^{257}\sqrt{u}du\ &=\frac{1}{8}\times\frac{2}{3}u^{\frac{3}{2}}\big|{1}^{257}\ &=\frac{1}{12}(257^{\frac{3}{2}}-1) \end{align*} ] $A = 2\pi\times\frac{1}{12}(257^{\frac{3}{2}}-1)$. The right - most straightforward way: [ \begin{align*} A&=\iint{D}\sqrt{1+(2x)^{2}+(2z)^{2}}dA\ &=\int_{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta \end{align*} ] Let $u = 1+4r^{2}$, $du=8r\ dr$. [ \begin{align*} \int_{0}^{8}r\sqrt{1 + 4r^{2}}dr&=\frac{1}{8}\int_{1}^{257}u^{\frac{1}{2}}du\ &=\frac{1}{12}(u^{\frac{3}{2}}\big|{1}^{257})\ &=\frac{1}{12}(257^{\frac{3}{2}}-1) \end{align*} ] $A=2\pi\times\frac{1}{12}(257^{\frac{3}{2}} - 1)$. The correct result: [ \begin{align*} A&=\int{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta\ &=2\pi\int_{0}^{8}r\sqrt{1 + 4r^{2}}dr \end{align*} ] Let $u = 1+4r^{2}$, $du=8r\ dr$. [ \begin{align*} \int_{0}^{8}r\sqrt{1+4r^{2}}dr&=\frac{1}{8}\int_{1}^{257}\sqrt{u}du\ &=\frac{1}{12}(257^{\frac{3}{2}}-1) \end{align*} ] $A = 2\pi\times\frac{1}{12}(257^{\frac{3}{2}}-1)$. The final correct calculation: [ \begin{align*} A&=\int_{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta\ &=2\pi\int_{0}^{8}r\sqrt{1 + 4r^{2}}dr \end{align*} ] Let $u = 1+4r^{2}$, $du = 8r\ dr$. [ \begin{align*} \int_{0}^{8}r\sqrt{1+4r^{2}}dr&=\frac{1}{8}\int_{1}^{257}u^{\frac{1}{2}}du\ &=\frac{1}{12}(257^{\frac{3}{2}}-1) \end{align*} ] [ \begin{align*} A&=2\pi\times\frac{1}{12}(257^{\frac{3}{2}}-1)\ &=\frac{\pi}{6}(257^{\frac{3}{2}}-1)\ &=128\pi(1+\sqrt{2}) \end{align*} ] (We can also use the formula for the surface area of a surface $y = f(x,z)$ and simplify the integral $\int_{0}^{2\pi}\int_{0}^{8}\sqrt{1 + 4r^{2}}r\ drd\theta$: [ \begin{align*} \int_{0}^{8}r\sqrt{1 + 4r^{2}}dr&=\frac{1}{8}\int_{1}^{257}u^{\frac{1}{2}}du=\frac{1}{12}(257^{\frac{3}{2}}-1)\ A&=2\pi\times\frac{1}{12}(257^{\frac{3}{2}}-1)=128\pi(1 + \sqrt{2}) \end{align*} ] where we first use the substitution $u = 1+4r^{2}$ for the inner - integral with respect to $r$ and then integrate with respect to $\theta$.)