Multivariate Calculus: Learning Objective 4

The region of integration is a rectangle and the integrand is a continuous function. You can apply Fubini's Theorem to set up an iterated integral. The order of integration (dx dy or dy dx) does not matter in this case.

$-11$

We set up the double integral as an iterated integral. Let's choose to integrate with respect to $x$ first, then $y$. $$ \iint\limits_R (xy-3xy^2)dA = \int_{1}^{2} \int_{0}^{2} (xy - 3xy^2) \, dx \, dy $$ First, we evaluate the inner integral with respect to $x$: $$ \int_{0}^{2} (xy - 3xy^2) \, dx = \left[ \frac{1}{2}x^2y - \frac{3}{2}x^2y^2 \right]_{x=0}^{x=2} $$ $$ = \left( \frac{1}{2}(2)^2y - \frac{3}{2}(2)^2y^2 \right) - (0) = 2y - 6y^2 $$ Now, we integrate this result with respect to $y$: $$ \int_{1}^{2} (2y - 6y^2) \, dy = \left[ y^2 - 2y^3 \right]_{1}^{2} $$ $$ = (2^2 - 2(2)^3) - (1^2 - 2(1)^3) = (4 - 16) - (1 - 2) = -12 - (-1) = -11 $$

Use the property that if $m \leq f(x,y) \leq M$ over a region $R$, then $m \cdot \text{Area}(R) \leq \iint\limits_R f(x,y)dA \leq M \cdot \text{Area}(R)$. You'll need to calculate the area of the rectangular region $R$.

$4 \leq \iint\limits_R (x^2+y^2)dA \leq 26$

The region $R$ is a rectangle defined by $1 \leq x \leq 3$ and $1 \leq y \leq 2$.

1. Calculate the area of R: $$ \text{Area}(R) = (\text{width}) \times (\text{height}) = (3-1) \times (2-1) = 2 \times 1 = 2 $$ 2. Identify the bounds of the function: The problem gives the lower bound $m=2$ and the upper bound $M=13$ for the function $f(x,y) = x^2+y^2$ on the region $R$.

3. Apply the bounding property: The lower bound for the integral is: $$ m \cdot \text{Area}(R) = 2 \cdot 2 = 4 $$ The upper bound for the integral is: $$ M \cdot \text{Area}(R) = 13 \cdot 2 = 26 $$ Thus, we have the inequality: $$ 4 \leq \iint\limits_R (x^2+y^2)dA \leq 26 $$

The integrand is a product of a function of only $x$ and a function of only $y$, and the integration region is a rectangle. This allows you to separate the double integral into a product of two independent single integrals.

$e-1$

Since the integrand $f(x,y) = e^y \cos(x)$ is a product of a function of $x$ and a function of $y$, and the region $R$ is a rectangle, we can separate the double integral: $$ \iint\limits_R e^y\cos(x)dA = \left( \int_{0}^{1} e^y \, dy \right) \left( \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx \right) $$ First, evaluate the integral with respect to $y$: $$ \int_{0}^{1} e^y \, dy = [e^y]_{0}^{1} = e^1 - e^0 = e - 1 $$ Next, evaluate the integral with respect to $x$: $$ \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx = [\sin(x)]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 $$ Finally, multiply the results: $$ (e-1) \times 1 = e-1 $$

The area of any region $R$ in the plane can be calculated by evaluating the double integral of the function $f(x,y) = 1$ over that region, i.e., Area = $\iint\limits_R 1 \, dA$.

$6$

To find the area, we set up the double integral of $f(x,y)=1$ over the rectangular region $R$. $$ \text{Area} = \iint\limits_R 1 \, dA = \int_{0}^{2} \int_{0}^{3} 1 \, dx \, dy $$ First, integrate with respect to $x$: $$ \int_{0}^{3} 1 \, dx = [x]_{0}^{3} = 3 - 0 = 3 $$ Now, integrate the result with respect to $y$: $$ \int_{0}^{2} 3 \, dy = [3y]_{0}^{2} = 3(2) - 3(0) = 6 $$ The area of the region is 6 square units.

The region $D$ is described as a Type I region (y-bounds depend on x). Therefore, you must integrate with respect to $y$ first. Notice that when integrating $e^{xy}$ with respect to $y$, $x$ is treated as a constant.

$2$

We set up the iterated integral according to the given region description: $$ \int_{0}^{2} \int_{x/2}^{1} x^2e^{xy} \, dy \, dx $$ First, integrate with respect to $y$: $$ \int_{x/2}^{1} x^2e^{xy} \, dy = x^2 \int_{x/2}^{1} e^{xy} \, dy = x^2 \left[ \frac{1}{x}e^{xy} \right]_{y=x/2}^{y=1} = x \left[ e^{xy} \right]_{y=x/2}^{y=1} $$ Now, evaluate at the bounds for $y$: $$ = x (e^{x(1)} - e^{x(x/2)}) = x(e^x - e^{x^2/2}) = xe^x - xe^{x^2/2} $$ Next, integrate this result with respect to $x$: $$ \int_{0}^{2} (xe^x - xe^{x^2/2}) \, dx = \int_{0}^{2} xe^x \, dx - \int_{0}^{2} xe^{x^2/2} \, dx $$ The first integral requires integration by parts ($\int u dv = uv - \int v du$ with $u=x, dv=e^x dx$): $$ \int_{0}^{2} xe^x \, dx = [xe^x - e^x]_{0}^{2} = (2e^2 - e^2) - (0 - e^0) = e^2 + 1 $$ The second integral is solved with a u-substitution ($u=x^2/2, du=x dx$): $$ \int_{0}^{2} xe^{x^2/2} \, dx = \int_{0}^{2} e^u \, du = [e^{x^2/2}]_{0}^{2} = e^{4/2} - e^0 = e^2 - 1 $$ Finally, subtract the results of the two integrals: $$ (e^2 + 1) - (e^2 - 1) = 2 $$

The iterated integral is already set up for you. Proceed by first evaluating the inner integral with respect to $v$. Be careful with the algebraic expansion after substituting the limits of integration.

$-63$

First, we integrate with respect to $v$: $$ \int_{-u^2-1}^{-u} 8uv \, dv = 8u \left[ \frac{1}{2}v^2 \right]_{-u^2-1}^{-u} = 4u [v^2]_{-u^2-1}^{-u} $$ Substitute the limits for $v$: $$ = 4u \left( (-u)^2 - (-u^2-1)^2 \right) = 4u \left( u^2 - (u^4 + 2u^2 + 1) \right) $$ $$ = 4u (u^2 - u^4 - 2u^2 - 1) = 4u (-u^4 - u^2 - 1) = -4u^5 - 4u^3 - 4u $$ Now, integrate this resulting polynomial with respect to $u$: $$ \int_{1}^{2} (-4u^5 - 4u^3 - 4u) \, du = \left[ -4\frac{u^6}{6} - 4\frac{u^4}{4} - 4\frac{u^2}{2} \right]_{1}^{2} $$ $$ = \left[ -\frac{2}{3}u^6 - u^4 - 2u^2 \right]_{1}^{2} $$ Evaluate at the limits for $u$: $$ = \left( -\frac{2}{3}(2)^6 - (2)^4 - 2(2)^2 \right) - \left( -\frac{2}{3}(1)^6 - (1)^4 - 2(1)^2 \right) $$ $$ = \left( -\frac{2}{3}(64) - 16 - 8 \right) - \left( -\frac{2}{3} - 1 - 2 \right) $$ $$ = \left( -\frac{128}{3} - 24 \right) - \left( -\frac{2}{3} - 3 \right) = \left( -\frac{128}{3} - \frac{72}{3} \right) - \left( -\frac{2}{3} - \frac{9}{3} \right) $$ $$ = -\frac{200}{3} - \left( -\frac{11}{3} \right) = -\frac{200}{3} + \frac{11}{3} = -\frac{189}{3} = -63 $$

First, integrate the polynomial $x+y^2$ with respect to $y$, treating $x$ as a constant. The limits for this inner integral are functions of $x$. Then integrate the resulting expression in $x$ from 0 to 1.

$\frac{23}{12}$

We begin with the inner integral with respect to $y$: $$ \int_{2x}^{3x} (x+y^2) \, dy = \left[ xy + \frac{1}{3}y^3 \right]_{y=2x}^{y=3x} $$ Substitute the limits of integration for $y$: $$ = \left( x(3x) + \frac{1}{3}(3x)^3 \right) - \left( x(2x) + \frac{1}{3}(2x)^3 \right) $$ $$ = \left( 3x^2 + \frac{27x^3}{3} \right) - \left( 2x^2 + \frac{8x^3}{3} \right) = (3x^2 + 9x^3) - (2x^2 + \frac{8}{3}x^3) $$ Combine like terms: $$ = (3-2)x^2 + (9 - \frac{8}{3})x^3 = x^2 + (\frac{27-8}{3})x^3 = x^2 + \frac{19}{3}x^3 $$ Now, perform the outer integration with respect to $x$: $$ \int_{0}^{1} \left( x^2 + \frac{19}{3}x^3 \right) \, dx = \left[ \frac{1}{3}x^3 + \frac{19}{3} \cdot \frac{1}{4}x^4 \right]_{0}^{1} = \left[ \frac{1}{3}x^3 + \frac{19}{12}x^4 \right]_{0}^{1} $$ Evaluate at the limits for $x$: $$ = \left( \frac{1}{3}(1)^3 + \frac{19}{12}(1)^4 \right) - (0) = \frac{1}{3} + \frac{19}{12} = \frac{4}{12} + \frac{19}{12} = \frac{23}{12} $$

The integration region is a rectangular box and the integrand is a product of functions of each variable ($f(x,y,z) = g(x)h(y)k(z)$). This allows the triple integral to be separated into a product of three single-variable integrals.

$162$

Because the integral is over a rectangular box and the integrand is separable, we can write: $$ \iiint\limits_B x^2yz \, dV = \left( \int_{-2}^{1} x^2 \, dx \right) \left( \int_{0}^{3} y \, dy \right) \left( \int_{1}^{5} z \, dz \right) $$ We evaluate each single integral separately.

The integral in $x$: $$ \int_{-2}^{1} x^2 \, dx = \left[ \frac{1}{3}x^3 \right]_{-2}^{1} = \frac{1}{3}(1)^3 - \frac{1}{3}(-2)^3 = \frac{1}{3} - \left(-\frac{8}{3}\right) = \frac{9}{3} = 3 $$ The integral in $y$: $$ \int_{0}^{3} y \, dy = \left[ \frac{1}{2}y^2 \right]_{0}^{3} = \frac{1}{2}(3)^2 - 0 = \frac{9}{2} $$ The integral in $z$: $$ \int_{1}^{5} z \, dz = \left[ \frac{1}{2}z^2 \right]_{1}^{5} = \frac{1}{2}(5)^2 - \frac{1}{2}(1)^2 = \frac{25}{2} - \frac{1}{2} = \frac{24}{2} = 12 $$ Finally, we multiply the results of the three integrals: $$ \text{Value} = 3 \times \frac{9}{2} \times 12 = 3 \times 9 \times 6 = 162 $$

First, set up the limits of integration for the tetrahedron. A common approach is to project the solid onto the $xy$-plane. This projection is the triangle bounded by $x=0$, $y=0$, and $x+y=1$. The limits will be $0 \le x \le 1$, $0 \le y \le 1-x$, and $0 \le z \le 1-x-y$.

$\frac{1}{12}$

We set up the integral based on the determined limits: $$ \iiint_T (5x-3y) \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} (5x-3y) \, dz \, dy \, dx $$ Since the integrand does not depend on $z$, the first integration is straightforward: $$ \int_{0}^{1-x-y} (5x-3y) \, dz = (5x-3y)[z]_{0}^{1-x-y} = (5x-3y)(1-x-y) $$ Now we have a double integral: $$ \int_{0}^{1} \int_{0}^{1-x} (5x-3y)(1-x-y) \, dy \, dx $$ Expand the integrand: $5x-5x^2-5xy-3y+3xy+3y^2 = 5x-5x^2-3y-2xy+3y^2$.
Integrate with respect to $y$: $$ \int_{0}^{1-x} (5x-5x^2-3y-2xy+3y^2) \, dy = \left[ (5x-5x^2)y - \frac{3}{2}y^2 - xy^2 + y^3 \right]_{0}^{1-x} $$ $$ = (5x-5x^2)(1-x) - \frac{3}{2}(1-x)^2 - x(1-x)^2 + (1-x)^3 $$ Let $u=1-x$. The expression becomes $(5x-5x^2)u - \frac{3}{2}u^2 - xu^2 + u^3 = 5x(1-x)u - \frac{3}{2}u^2 - xu^2 + u^3 = 5xu^2 - \frac{3}{2}u^2 - xu^2 + u^3 = (4x - \frac{3}{2})u^2 + u^3$. Substitute $u=1-x$ back: $(4x-\frac{3}{2})(1-x)^2 + (1-x)^3$. $$ = (4x-\frac{3}{2})(1-2x+x^2) + (1-3x+3x^2-x^3) $$ $$ = (4x-8x^2+4x^3-\frac{3}{2}+3x-\frac{3}{2}x^2) + (1-3x+3x^2-x^3) $$ $$ = 3x^3 + (-\frac{13}{2})x^2 + 4x - \frac{1}{2} $$ Finally, integrate with respect to $x$: $$ \int_{0}^{1} (3x^3 - \frac{13}{2}x^2 + 4x - \frac{1}{2}) \, dx = \left[ \frac{3}{4}x^4 - \frac{13}{6}x^3 + 2x^2 - \frac{1}{2}x \right]_{0}^{1} $$ $$ = \frac{3}{4} - \frac{13}{6} + 2 - \frac{1}{2} = \frac{9-26+24-6}{12} = \frac{1}{12} $$

The volume can be found using the formula $V = \frac{1}{3} \times (\text{Base Area}) \times (\text{Height})$. Alternatively, you can use integration by slicing. Consider a horizontal cross-section (a square) at an arbitrary height $z$. Find the area of this square, $A(z)$, and then integrate $A(z)$ from the base to the vertex.

$\frac{4}{3}$

We can solve this using the method of slicing. The pyramid has a height of $H=1$. The base is a square with side length 2, so its area is $A_{base} = 2 \times 2 = 4$.

Let's consider a horizontal cross-section at a height $z$ from the $xy$-plane. This cross-section is also a square. Let its side length be $s$. We can use similar triangles to relate the side length $s$ to the height $z$.

Consider the triangle formed by the pyramid's vertex $(0,0,1)$, the center of the base $(0,0,0)$, and the midpoint of a base edge, e.g., $(1,0,0)$. The height of this triangle is 1 and its base is 1. The cross-section at height $z$ creates a smaller, similar triangle with height $1-z$ and base $s/2$.

By similar triangles: $$ \frac{\text{base of small triangle}}{\text{height of small triangle}} = \frac{\text{base of large triangle}}{\text{height of large triangle}} $$ $$ \frac{s/2}{1-z} = \frac{1}{1} \implies s = 2(1-z) $$ The area of the square cross-section at height $z$ is $A(z) = s^2 = [2(1-z)]^2 = 4(1-z)^2$.

The volume of the pyramid is the integral of these cross-sectional areas from $z=0$ to $z=1$: $$ V = \int_{0}^{1} A(z) \, dz = \int_{0}^{1} 4(1-z)^2 \, dz $$ We can use a u-substitution with $u=1-z$, so $du = -dz$. When $z=0$, $u=1$. When $z=1$, $u=0$. $$ V = \int_{1}^{0} 4u^2 (-du) = -4 \int_{1}^{0} u^2 \, du = 4 \int_{0}^{1} u^2 \, du $$ $$ V = 4 \left[ \frac{u^3}{3} \right]_{0}^{1} = 4 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = \frac{4}{3} $$ The volume is $\frac{4}{3}$ cubic units.