15.1: Double Integrals over Rectangles #
Problem 1: Define the solid that lies below the surface $f(x, y) = xy$ and above the rectangle
$$ R = {(x, y) : 0 \leq x \leq 6, 0 \leq y \leq 4 } $$
$(a)$ Estimate the volume by using a Riemann sum with $m = 3$, $n = 2$, where the sample point is the upper right corner of each square.
Being pedantic, we break up our rectangle $R$ into $6$ squares,
$$ [0, 2] \times [0, 2], [2, 4] \times [0, 2], [4, 6] \times [0, 2] $$
$$ [0, 2] \times [2, 4], [2, 4] \times [2, 4], [4, 6] \times [2, 4] $$
Each of these squares has volume $4$, so our estimation becomes,
$$ 4 (f(2, 2) + f(4, 2) + f(6, 2) + f(2, 4) + f(4, 4) + f(6, 4)) $$
$$ = 4 (4 + 8 + 12 + 8 + 16 + 24) = 288 $$
$(b)$ Use the midpoint rule to estimate the volume of the solid.
We already have our squares, so we just sum over the midpoints,
$$ 4 (f(1, 1) + f(3, 1) + f(5, 1) + f(1, 3) + f(3, 3) + f(5, 3)) $$
$$ = 4 (1 + 3 + 5 + 3 + 9 + 15) = 144 $$
Problem 9: Evaluate the double integral by first identifying it as the volume of a solid.
$$ \iint_R \sqrt{2} \ dA $$
where $R = {(x, y) : 2 \leq x \leq 6, -1 \leq y \leq 5 }$.
Note that since this function is constant, the integral is just the volume of the rectangular prism that has a base of $R$ and the hight of $\sqrt{2}$. Thus the integral evaluates to
$$ \sqrt{2} (6 - 2) (5 - (-1)) = 24 \sqrt{2} $$
Problem 15: Calculate the iterated integral,
$$ \int^4_1 \int^2_0 (6x^2y - 2x) \ dy \ dx $$
First we evaluate the inner integral,
$$ \int^2_0 (6x^2y - 2x) \ dy = (3x^2y^2 - 2xy) \big\vert^2_0 = 12x^2 - 4x $$
Then the next integral is simple,
$$ \int^4_1 (12x^2 - 4x) \ dx = (4x^3 - 2x^2) \big\vert^4_1 = 256 - 32 - 4 + 2 = 222 $$
Problem 27: Calculate the double integral,
$$ \iint_R x \sec^2(y) \ dA $$
where $R = {(x, y) : 0 \leq x \leq 2, 0 \leq y \leq \pi/4}$.
Fubini’s theorem allows us to convert this to an interative integral, so we just integrate the following iterative integral,
$$ \iint_R x \sec^2(y) \ dA = \int^2_0 \int^{\pi/4}_0 x \sec^2(y) \ dy \ dx = \int^2_0 x (\tan(\pi/4) - \tan(0)) \ dx = 2 $$
15.2: Double Integrals over General Regions #
Problem 3: Calculate the iterated integral,
$$ \int^1_0 \int^y_0 xe^{y^3} \ dx \ dy $$
The inner integral is not so bad, it comes out to,
$$ \int^y_0 xe^{y^3} \ dx = \frac{x^2e^{y^3}}{2} \big\vert^y_0 = \frac{y^2e^{y^3}}{2} $$
Now we do a $u$-sub where $u = y^3$, then $du/3 = y^2dy$. So we have,
$$ \int^1_0 \frac{y^2e^{y^3}}{2} \ dy = \int^1_0 \frac{e^u}{6} \ du = \frac{e - 1}{6} $$
Problem 10: Calculate the double integral,
$$ \iint_D y \sqrt{x^2 - y^2} \ dA $$
where $D = {(x, y) : 0 \leq x \leq 2, 0 \leq y \leq x}$.
Using Fubini’s theorem, we may turn this into an iterated integral,
$$ \iint_D y \sqrt{x^2 - y^2} \ dA = \int^2_0 \int^x_0 y \sqrt{x^2 - y^2} \ dy \ dx $$
We do a $u$-substitution, setting $u = x^2 - y^2$ thus, $du = -2ydy$, then we have
$$ \int^2_0 \int^x_0 y \sqrt{x^2 - y^2} \ dy \ dx = \int^2_0 \int^b_a - \frac{\sqrt{u}}{2} \ du \ dx $$
$$ = \int^2_0 - \frac{2(x^2 - y^2)^{3/2}}{6} \bigg \vert^{y=x}_{y=0} \ dx = \int^2_0 \frac{2|x|^3}{6} \ dx = \frac{4}{3} $$
Problem 30: Find the volume of the solid that is bounded by the cylinder $y^2 + z^2 = 4$ and the planes $x = 2y$, $x = 0$, $z = 0$ in the first octant.