Let's solve both problems step-by-step. --- ### **Problem 2** **Find the volume under the surface \( z = \sqrt{x^2 + y^2} \) above the triangular region bounded by \( x=0 \), \( y=0 \), and \( x+y=1 \).** #### **Step 1: Sketch and Set Up the Region** The region \( R \) is bounded by: - \( x = 0 \) (y-axis) - \( y = 0 \) (x-axis) - \( x + y = 1 \) (line from \((1,0)\) to \((0,1)\)) For a fixed \( x \), \( y \) ranges from \( 0 \) to \( 1-x \). \( x \) ranges from \( 0 \) to \( 1 \). #### **Step 2: Set Up the Double Integral** The volume is: \[ V = \iint_{R} \sqrt{x^2 + y^2} \, dA \] In terms of \( x \) and \( y \): \[ V = \int_{x=0}^{1} \int_{y=0}^{1-x} \sqrt{x^2 + y^2} \, dy \, dx \] #### **Step 3: Integrate with Respect to \( y \)** Let’s compute the inner integral: \[ I = \int_{0}^{1-x} \sqrt{x^2 + y^2} \, dy \] Let \( y = t \), \( x \) is a constant. Recall: \[ \int \sqrt{a^2 + t^2} \, dt = \frac{t}{2} \sqrt{a^2 + t^2} + \frac{a^2}{2} \ln \left( t + \sqrt{a^2 + t^2} \right) + C \] With \( a = x \): \[ \int_{0}^{1-x} \sqrt{x^2 + y^2} \, dy = \left[ \frac{y}{2} \sqrt{x^2 + y^2} + \frac{x^2}{2} \ln \left( y + \sqrt{x^2 + y^2} \right) \right]_{y=0}^{y=1-x} \] Plug in the limits: - At \( y = 1-x \): - \( \sqrt{x^2 + (1-x)^2} = \sqrt{x^2 + 1 - 2x + x^2} = \sqrt{2x^2 - 2x + 1} \) - \( y + \sqrt{x^2 + y^2} = (1-x) + \sqrt{2x^2 - 2x + 1} \) - At \( y = 0 \): - \( \sqrt{x^2 + 0} = x \) - \( 0 + x = x \) So, \[ I = \frac{1-x}{2} \sqrt{2x^2 - 2x + 1} + \frac{x^2}{2} \ln \left[ (1-x) + \sqrt{2x^2 - 2x + 1} \right] - \left( 0 + \frac{x^2}{2} \ln x \right) \] \[ = \frac{1-x}{2} \sqrt{2x^2 - 2x + 1} + \frac{x^2}{2} \ln \left( \frac{(1-x) + \sqrt{2x^2 - 2x + 1}}{x} \right) \] #### **Step 4: Integrate with Respect to \( x \)** The final answer is: \[ V = \int_{0}^{1} \left\{ \frac{1-x}{2} \sqrt{2x^2 - 2x + 1} + \frac{x^2}{2} \ln \left( \frac{(1-x) + \sqrt{2x^2 - 2x + 1}}{x} \right) \right\} dx \] **This integral can be evaluated numerically, but the above is the exact answer in terms of elementary functions.** --- ### **Problem 3** **Evaluate the triple integral:** \[ \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{1-x^2-y^2} (1 - z^2) \, dz \, dy \, dx \] #### **Step 1: Integrate with Respect to \( z \) (Innermost Integral)** \[ \int_{0}^{1-x^2-y^2} (1-z^2) \, dz \] Let’s integrate: \[ = \left[ z - \frac{z^3}{3} \right]_{z=0}^{z=1-x^2-y^2} \] \[ = (1-x^2-y^2) - \frac{(1-x^2-y^2)^3}{3} \] #### **Step 2: Substitute Back** So, the triple integral becomes: \[ \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left[ (1-x^2-y^2) - \frac{(1-x^2-y^2)^3}{3} \right] dy\, dx \] #### **Step 3: Recognize the Region** - The region in the \( xy \)-plane is \( x^2 + y^2 \leq 1 \) (a unit disk). #### **Step 4: Switch to Polar Coordinates** Let \( x = r \cos \theta \), \( y = r \sin \theta \), where \( r \) from \( 0 \) to \( 1 \), \( \theta \) from \( 0 \) to \( 2\pi \), and \( x^2 + y^2 = r^2 \). The Jacobian is \( r \). So, the integral becomes: \[ \int_{0}^{2\pi} \int_{0}^{1} \left[ (1 - r^2) - \frac{(1 - r^2)^3}{3} \right] r \, dr\, d\theta \] #### **Step 5: Simplify and Integrate** First, expand: \[ (1 - r^2) - \frac{(1 - r^2)^3}{3} \] Calculate \( (1 - r^2)^3 = 1 - 3r^2 + 3r^4 - r^6 \): \[ = (1 - r^2) - \frac{1}{3} (1 - 3r^2 + 3r^4 - r^6) \] \[ = 1 - r^2 - \frac{1}{3} + r^2 - r^4 + \frac{1}{3} r^6 \] \[ = \left(1 - \frac{1}{3}\right) + ( - r^2 + r^2) - r^4 + \frac{1}{3} r^6 \] \[ = \frac{2}{3} - r^4 + \frac{1}{3} r^6 \] So the integrand is: \[ \left( \frac{2}{3} - r^4 + \frac{1}{3} r^6 \right) r \] So the double integral is: \[ \int_{0}^{2\pi} \int_{0}^{1} \left( \frac{2}{3} r - r^5 + \frac{1}{3} r^7 \right) dr\, d\theta \] #### **Step 6: Integrate with Respect to \( r \)** \[ \int_{0}^{1} \frac{2}{3} r dr = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} \] \[ \int_{0}^{1} r^5 dr = \frac{1}{6} \] \[ \int_{0}^{1} r^7 dr = \frac{1}{8} \] So, \[ \int_{0}^{1} \left( \frac{2}{3} r - r^5 + \frac{1}{3} r^7 \right) dr = \frac{1}{3} - \frac{1}{6} + \frac{1}{3} \cdot \frac{1}{8} = \frac{1}{3} - \frac{1}{6} + \frac{1}{24} \] Find common denominator (24): \[ = \frac{8}{24} - \frac{4}{24} + \frac{1}{24} = \frac{5}{24} \] #### **Step 7: Integrate with Respect to \( \theta \)** \[ \int_{0}^{2\pi} d\theta = 2\pi \] So, the final answer: \[ 2\pi \cdot \frac{5}{24} = \frac{5\pi}{12} \] --- ## **Final Answers** ### 2. The volume under the surface is: \[ V = \int_{0}^{1} \left\{ \frac{1-x}{2} \sqrt{2x^2-2x+1} + \frac{x^2}{2} \ln \left( \frac{(1-x) + \sqrt{2x^2-2x+1}}{x} \right) \right\} dx \] ### 3. The value of the triple integral is: \[ \boxed{\frac{5\pi}{12}} \]

Let's carefully go through each problem step-by-step with detailed calculations. --- ## **Problem 2: Find the volume under the surface \( z = \sqrt{x^2 + y^2} \) over the triangular region bounded by \( x=0 \), \( y=0 \), and \( x + y = 1 \).** --- ### **Step 1: Understand the region \( R \)** - The region \( R \) is in the first quadrant, bounded by: - \( x=0 \) (the y-axis), - \( y=0 \) (the x-axis), - \( x + y = 1 \) (a line from \((1,0)\) to \((0,1)\)). - For fixed \( x \in [0,1] \), \( y \) varies from \( 0 \) to \( 1 - x \). --- ### **Step 2: Set up the double integral** The volume \( V \) is: \[ V = \iint_{R} z \, dA = \iint_{R} \sqrt{x^2 + y^2}\, dy\, dx \] with limits: - \( x \in [0, 1] \) - \( y \in [0, 1 - x] \) Thus, \[ V = \int_{x=0}^{1} \int_{y=0}^{1 - x} \sqrt{x^2 + y^2}\, dy\, dx \] --- ### **Step 3: Inner integral with respect to \( y \)** Treat \( x \) as a constant: \[ I(x) = \int_{0}^{1 - x} \sqrt{x^2 + y^2}\, dy \] **Standard integral:** \[ \int \sqrt{a^2 + t^2} \, dt = \frac{t}{2} \sqrt{a^2 + t^2} + \frac{a^2}{2} \ln \left| t + \sqrt{a^2 + t^2} \right| + C \] Set \( a = x \), \( t = y \): \[ I(x) = \left[ \frac{y}{2} \sqrt{x^2 + y^2} + \frac{x^2}{2} \ln \left| y + \sqrt{x^2 + y^2} \right| \right]_0^{1 - x} \] **Evaluate at the limits:** - At \( y = 1 - x \): \[ \sqrt{x^2 + (1 - x)^2} = \sqrt{x^2 + 1 - 2x + x^2} = \sqrt{2x^2 - 2x + 1} \] \[ \Rightarrow \text{Expression} = \frac{1 - x}{2} \sqrt{2x^2 - 2x + 1} + \frac{x^2}{2} \ln \left( (1 - x) + \sqrt{2x^2 - 2x + 1} \right) \] - At \( y=0 \): \[ \sqrt{x^2 + 0} = x \] \[ \Rightarrow \text{Expression} = 0 + \frac{x^2}{2} \ln x \] **Therefore,** \[ I(x) = \frac{1 - x}{2} \sqrt{2x^2 - 2x + 1} + \frac{x^2}{2} \ln \left( (1 - x) + \sqrt{2x^2 - 2x + 1} \right) - \frac{x^2}{2} \ln x \] --- ### **Step 4: Final integral over \( x \)** The volume: \[ V = \int_0^1 I(x)\, dx \] \[ V = \int_0^1 \left[ \frac{1 - x}{2} \sqrt{2x^2 - 2x + 1} + \frac{x^2}{2} \ln \left( \frac{(1 - x) + \sqrt{2x^2 - 2x + 1}}{x} \right) \right] dx \] **This is an exact expression for the volume.** --- ## **Problem 3: Evaluate the triple integral** \[ \iiint_{D} (1 - z^2) \, dz\, dy\, dx \] where \( D \) is bounded by: - \( x^2 + y^2 \le 1 \) (a unit disk in the xy-plane), - \( 0 \le z \le 1 - x^2 - y^2 \). --- ### **Step 1: Recognize the region** - The projection onto the xy-plane is the unit disk \( r \in [0,1] \), \( \theta \in [0, 2\pi] \). - For each point \( (x,y) \), \( z \) varies from \( 0 \) up to \( 1 - x^2 - y^2 \). ### **Step 2: Switch to polar coordinates** Set: \[ x = r \cos \theta,\quad y = r \sin \theta,\quad z \in [0, 1 - r^2] \] Jacobian: \[ dx\, dy = r\, dr\, d\theta \] The integral becomes: \[ I = \int_{\theta=0}^{2\pi} \int_{r=0}^{1} \left[ \int_{z=0}^{1 - r^2} (1 - z^2) dz \right] r\, dr\, d\theta \] ### **Step 3: Inner integral over \( z \)** \[ J(r) = \int_{0}^{1 - r^2} (1 - z^2) dz \] Compute: \[ J(r) = \left[ z - \frac{z^3}{3} \right]_0^{1 - r^2} = (1 - r^2) - \frac{(1 - r^2)^3}{3} \] ### **Step 4: The integral reduces to** \[ I = \int_{0}^{2\pi} \int_{0}^{1} \left[ (1 - r^2) - \frac{(1 - r^2)^3}{3} \right] r\, dr\, d\theta \] Since the integrand is independent of \( \theta \): \[ I = 2\pi \int_0^1 \left[ (1 - r^2) - \frac{(1 - r^2)^3}{3} \right] r\, dr \] --- ### **Step 5: Simplify the integrand** Let’s expand \( (1 - r^2)^3 \): \[ (1 - r^2)^3 = 1 - 3r^2 + 3r^4 - r^6 \] Substitute into the integral: \[ I_r = \int_0^1 \left[ (1 - r^2) - \frac{1 - 3r^2 + 3r^4 - r^6}{3} \right] r\, dr \] Simplify numerator: \[ (1 - r^2) - \frac{1}{3} + r^2 - r^4 + \frac{r^6}{3} \] Expressed as: \[ \left( 1 - \frac{1}{3} \right) + (- r^2 + r^2) - r^4 + \frac{r^6}{3} \] which simplifies to: \[ \frac{2}{3} - r^4 + \frac{r^6}{3} \] Now multiply by \( r \): \[ \left( \frac{2}{3} - r^4 + \frac{r^6}{3} \right) r = \frac{2}{3} r - r^5 + \frac{r^7}{3} \] --- ### **Step 6: Integrate with respect to \( r \)** \[ I_r = \int_0^1 \left( \frac{2}{3} r - r^5 + \frac{r^7}{3} \right) dr \] Calculate term-by-term: 1. \( \int_0^1 \frac{2}{3} r\, dr = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} \) 2. \( \int_0^1 r^5\, dr = \frac{1}{6} \) 3. \( \int_0^1 \frac{r^7}{3}\, dr = \frac{1}{3} \cdot \frac{1}{8} = \frac{1}{24} \) Now sum: \[ I_r = \frac{1}{3} - \frac{1}{6} + \frac{1}{24} \] Find common denominator 24: \[ = \frac{8}{24} - \frac{4}{24} + \frac{1}{24} = \frac{8 - 4 + 1}{24} = \frac{5}{24} \] ### **Step 7: Multiply by \( 2\pi \) for \( \theta \) integral** \[ I = 2\pi \times \frac{5}{24} = \frac{10\pi}{24} = \frac{5\pi}{12} \] --- ## **Final answers:** ### **Problem 2:** \[ \boxed{ V = \int_0^1 \left[ \frac{1 - x}{2} \sqrt{2x^2 - 2x + 1} + \frac{x^2}{2} \ln \left( \frac{(1 - x) + \sqrt{2x^2 - 2x + 1}}{x} \right) \right] dx } \] *(Exact expression; numerical approximation can be computed if needed)* ### **Problem 3:** \[ \boxed{ \frac{5\pi}{12} } \] --- **This completes the detailed step-by-step solution.**