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MTH 254 Calculus IV Winter 2026

Richard Zhao

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Worksheet Assigned Problems for Section 15.1

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.
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The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.
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15.1.1.

Compute the Riemann sum \(S_{4,3}\) to estimate the double integral of \(f(x,y) = xy\) over \(\c{R} = [1,3] \times [1,2.5]\text{.}\) Use the regular partition and upper-right vertices of the sub-rectangles as sample points.
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Solution.
We are asked to compute \(S_{4,3}\) for \(f(x,y) = xy\) on the region \(\c{R} = [1,3] \times [1,2.5]\text{.}\) This means we partition the \(x\)-interval \([1,3]\) into \(N=4\) subintervals and the \(y\)-interval \([1,2.5]\) into \(M=3\) subintervals.
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First, let’s find the dimensions of our sub-rectangles:
\begin{align*} \Delta x \amp = \frac{3 - 1}{4} = \frac{2}{4} = 0.5 \\ \Delta y \amp = \frac{2.5 - 1}{3} = \frac{1.5}{3} = 0.5 \\ \Delta A \amp = \Delta x \Delta y = (0.5)(0.5) = 0.25 \end{align*}
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The \(x\)-grid points are: \(1, 1.5, 2, 2.5, 3\text{.}\) The \(y\)-grid points are: \(1, 1.5, 2, 2.5\text{.}\)
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We are using the upper-right vertices as our sample points \((x_i^*, y_j^*)\text{.}\) This means for each sub-rectangle, we take the largest \(x\) and the largest \(y\) value. The sample points will be combinations of \(x \in \{1.5, 2, 2.5, 3\}\) and \(y \in \{1.5, 2, 2.5\}\text{.}\)
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Since \(f(x,y) = xy\text{,}\) the Riemann sum is:
\begin{align*} S_{4,3} \amp = \sum_{i=1}^4 \sum_{j=1}^3 f(x_i^*, y_j^*) \Delta A \\ \amp = 0.25 \sum_{i=1}^4 \sum_{j=1}^3 (x_i^*)(y_j^*) \end{align*}
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Because the function factors nicely into \(f(x)g(y)\text{,}\) we can actually factor the summation:
\begin{align*} S_{4,3} \amp = 0.25 \left( \sum_{i=1}^4 x_i^* \right) \left( \sum_{j=1}^3 y_j^* \right) \\ \amp = 0.25 \big(1.5 + 2 + 2.5 + 3\big) \big(1.5 + 2 + 2.5\big) \\ \amp = 0.25 (9)(6) \\ \amp = 0.25 (54) = 13.5 \end{align*}
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The estimated double integral is \(13.5\text{.}\)
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15.1.9.

Evaluate \(\ds \iint_\c{R} (15 - 3x)\, dA\text{,}\) where \(\c{R} = [0,5] \times [0,3]\text{,}\) and sketch the corresponding solid region.
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Solution.
The surface is a plane \(z = 15 - 3x\text{.}\) Let’s look at the region \(\c{R} = [0,5] \times [0,3]\text{.}\)
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When \(x = 0\text{,}\) the height is \(z = 15 - 3(0) = 15\text{.}\) When \(x = 5\text{,}\) the height is \(z = 15 - 3(5) = 0\text{.}\) Notice that the height does not depend on \(y\text{.}\)
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Because \(z\) goes from \(15\) linearly down to \(0\) as \(x\) goes from \(0\) to \(5\text{,}\) the solid region under this surface is a triangular prism (or a wedge).
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We can calculate the volume using geometry. The base of the triangular face (in the \(xz\)-plane) has length 5 (from \(x=0\) to \(5\)), and the height is 15. The "length" of the prism stretches along the \(y\)-axis from \(y=0\) to \(3\text{.}\)
\begin{align*} \text{Area of triangular face} \amp = \frac{1}{2}bh = \frac{1}{2}(5)(15) = \frac{75}{2} = 37.5 \\ \text{Volume} \amp = \text{Area} \times \text{length} = 37.5 \times 3 = 112.5 \end{align*}
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Alternatively, we can evaluate it using an iterated integral:
\begin{align*} \int_0^3 \int_0^5 (15 - 3x)\, dx\, dy \amp = \int_0^3 \left[ 15x - \frac{3}{2}x^2 \right]_0^5\, dy \\ \amp = \int_0^3 \left( 15(5) - \frac{3}{2}(25) \right)\, dy \\ \amp = \int_0^3 \left( 75 - 37.5 \right)\, dy \\ \amp = \int_0^3 37.5\, dy = [37.5y]_0^3 = 112.5 \end{align*}
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15.1.11.

The following table gives the approximate height at quarter-meter intervals of a mound of gravel. Estimate the volume of the mound by computing the average of the two Riemann sums \(S_{4,3}\) with lower-left and upper-right vertices of the sub-rectangles as sample points.
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Solution.
Each subrectangle is a square of side \(0.25\text{,}\) hence the area of each subrectangle is \(\Delta A = 0.25^2 = 0.0625\text{.}\) By the given data, the lower-left vertex sample points are:
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\begin{align*} f(P_{11}) \amp= f(0, 0) \amp f(P_{12}) \amp= f(0, 0.25) \amp f(P_{13}) \amp= f(0, 0.50) \\ f(P_{21}) \amp= f(0.25, 0) \amp f(P_{22}) \amp= f(0.25, 0.25) \amp f(P_{23}) \amp= f(0.25, 0.50) \\ f(P_{31}) \amp= f(0.50, 0) \amp f(P_{32}) \amp= f(0.50, 0.25) \amp f(P_{33}) \amp= f(0.50, 0.50) \\ f(P_{41}) \amp= f(0.75, 0) \amp f(P_{42}) \amp= f(0.75, 0.25) \amp f(P_{43}) \amp= f(0.75, 0.50) \end{align*}
The Riemann sum \(S_{4,3}\) that corresponds to these lower-left vertex sample points is the following sum:
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\begin{align*} S_{4,3} \amp= \sum_{i=1}^4 \sum_{j=1}^3 f(P_{ij}) \Delta A \\ \amp= 0.0625(0.1 + 0.15 + 0.2 + 0.15 + 0.2 + 0.3 + 0.2 + 0.4 + 0.5 + 0.15 + 0.3 + 0.4) \\ \amp\approx 0.190625 \end{align*}
Now by the given data, the upper-right vertex sample points are:
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\begin{align*} f(P_{11}) \amp= f(0.25, 0.25) \amp f(P_{12}) \amp= f(0.25, 0.50) \amp f(P_{13}) \amp= f(0.25, 0.75) \\ f(P_{21}) \amp= f(0.50, 0.25) \amp f(P_{22}) \amp= f(0.50, 0.50) \amp f(P_{23}) \amp= f(0.50, 0.75) \\ f(P_{31}) \amp= f(0.75, 0.25) \amp f(P_{32}) \amp= f(0.75, 0.50) \amp f(P_{33}) \amp= f(0.75, 0.75) \\ f(P_{41}) \amp= f(1, 0.25) \amp f(P_{42}) \amp= f(1, 0.50) \amp f(P_{43}) \amp= f(1, 0.75) \end{align*}
The Riemann sum \(S'_{4,3}\) that corresponds to these upper-right vertex sample points is the following sum:
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\begin{align*} S'_{4,3} \amp= \sum_{i=1}^4 \sum_{j=1}^3 f(P_{ij}) \Delta A \\ \amp= 0.0625(0.2 + 0.3 + 0.2 + 0.4 + 0.5 + 0.2 + 0.3 + 0.4 + 0.15 + 0.2 + 0.2 + 0.1) \\ \amp\approx 0.196875 \end{align*}
Taking the average of the two Riemann sums we have:
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\begin{align*} \text{volume} \amp\approx \frac{S_{4,3} + S'_{4,3}}{2} = \frac{0.190625 + 0.196875}{2} = 0.19375 \end{align*}
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15.1.17.

Use symmetry to evaluate the double integral
\begin{equation*} \iint_\c{R} \sin(x)\, dA\, , \qquad \c{R} = [0,2\pi] \times [0,2\pi] \end{equation*}
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Solution.
We are asked to evaluate \(\ds \iint_\c{R} \sin(x)\, dA\) over the region \([0,2\pi] \times [0,2\pi]\) using symmetry.
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Notice that the integrand \(f(x,y) = \sin(x)\) is completely independent of \(y\text{.}\) This means the solid is just the sine wave extended along the \(y\)-axis from \(y=0\) to \(y=2\pi\) (like a corrugated metal roof).
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Let’s look at the cross-section of the surface along the \(x\)-axis. The function \(\sin(x)\) on the interval \([0, 2\pi]\) is perfectly symmetric about the \(x\)-axis. The "positive" volume above the \(xy\)-plane from \(x=0\) to \(\pi\) is exactly equal in magnitude to the "negative" volume below the \(xy\)-plane from \(x=\pi\) to \(2\pi\text{.}\)
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Because these two identical volumes have opposite signs, they cancel each other out completely. Therefore:
\begin{equation*} \iint_\c{R} \sin(x)\, dA = 0 \end{equation*}
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Exercise Group.

In the following exercises, evaluate the iterated integrals.
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15.1.23.
\(\ds \int_{-1}^1 \int_0^\pi x^2\sin(y)\, dy \, dx\)
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Solution.
We evaluate the inner integral first with respect to \(y\text{,}\) treating \(x\) as a constant:
\begin{align*} \int_0^\pi x^2\sin(y)\, dy \amp = x^2 \Big[ -\cos(y) \Big]_0^\pi \\ \amp = x^2 \big( -\cos(\pi) - (-\cos(0)) \big) \\ \amp = x^2 \big( -(-1) + 1 \big) = 2x^2 \end{align*}
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Now, we substitute this back into the outer integral and integrate with respect to \(x\text{:}\)
\begin{align*} \int_{-1}^1 2x^2\, dx \amp = \left[ \frac{2}{3}x^3 \right]_{-1}^1 \\ \amp = \frac{2}{3}(1)^3 - \frac{2}{3}(-1)^3 \\ \amp = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \end{align*}
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15.1.33.
\(\ds \int_0^4 \int_0^5 \frac{dy \, dx}{\sqrt{x + y}}\)
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Solution.
We rewrite the integrand using a fractional exponent: \(\frac{1}{\sqrt{x+y}} = (x+y)^{-1/2}\text{.}\) Evaluating the inner integral with respect to \(y\) (treating \(x\) as a constant):
\begin{align*} \int_0^5 (x + y)^{-1/2}\, dy \amp = \left[ 2(x + y)^{1/2} \right]_{y=0}^{y=5} \\ \amp = 2(x + 5)^{1/2} - 2(x + 0)^{1/2} \\ \amp = 2\sqrt{x + 5} - 2\sqrt{x} \end{align*}
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Now substitute this into the outer integral and integrate with respect to \(x\text{:}\)
\begin{align*} \int_0^4 \left( 2(x + 5)^{1/2} - 2x^{1/2} \right)\, dx \amp = \left[ 2\left(\frac{2}{3}\right)(x + 5)^{3/2} - 2\left(\frac{2}{3}\right)x^{3/2} \right]_0^4 \\ \amp = \frac{4}{3} \left[ (x + 5)^{3/2} - x^{3/2} \right]_0^4 \end{align*}
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Evaluate at the boundaries:
\begin{align*} \amp \frac{4}{3} \Big( \big( (4+5)^{3/2} - 4^{3/2} \big) - \big( (0+5)^{3/2} - 0^{3/2} \big) \Big) \\ \amp = \frac{4}{3} \Big( (9^{3/2} - 4^{3/2}) - (5^{3/2} - 0) \Big) \\ \amp = \frac{4}{3} \Big( (27 - 8) - 5\sqrt{5} \Big) \\ \amp = \frac{4}{3} (19 - 5\sqrt{5}) \end{align*}
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Exercise Group.

In the following exercises, evaluate the integral.
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15.1.39.
\(\ds \iint_\c{R} \cos(x)\sin(2y)\, dA \, , \qquad \c{R} = \left[0,\frac{\pi}{2}\right] \times \left[0,\frac{\pi}{2}\right]\)
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Solution.
We will evaluate this as an iterated integral. Let’s choose to integrate with respect to \(x\) first, treating \(y\) as a constant:
\begin{align*} \int_0^{\pi/2} \cos(x)\sin(2y)\, dx \amp = \sin(2y) \Big[ \sin(x) \Big]_{x=0}^{x=\pi/2} \\ \amp = \sin(2y) \left( \sin\left(\frac{\pi}{2}\right) - \sin(0) \right) \\ \amp = \sin(2y) (1 - 0) = \sin(2y) \end{align*}
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Next, we substitute this result into the outer integral and integrate with respect to \(y\text{.}\) We’ll use a small u-substitution (\(u = 2y, du = 2dy\)):
\begin{align*} \int_0^{\pi/2} \sin(2y)\, dy \amp = \left[ -\frac{1}{2}\cos(2y) \right]_0^{\pi/2} \\ \amp = -\frac{1}{2}\cos\left(2\cdot\frac{\pi}{2}\right) - \left(-\frac{1}{2}\cos(0)\right) \\ \amp = -\frac{1}{2}\cos(\pi) + \frac{1}{2}\cos(0) \\ \amp = -\frac{1}{2}(-1) + \frac{1}{2}(1) = \frac{1}{2} + \frac{1}{2} = 1 \end{align*}
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15.1.43.
\(\ds \iint_\c{R} x\ln(y)\, dA\, , \qquad \c{R} = [0,3] \times [1,e]\)
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Solution.
We will set this up as an iterated integral. Let’s integrate with respect to \(x\) first, keeping \(y\) constant:
\begin{align*} \int_0^3 x\ln(y)\, dx \amp = \ln(y) \left[ \frac{1}{2}x^2 \right]_{x=0}^{x=3} \\ \amp = \ln(y) \left( \frac{1}{2}(3)^2 - \frac{1}{2}(0)^2 \right) \\ \amp = \frac{9}{2}\ln(y) \end{align*}
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Now, we substitute this back into the outer integral and integrate with respect to \(y\text{.}\) Recall from MTH 252Z that integrating \(\ln(y)\) requires Integration by Parts (\(u = \ln(y), dv = dy\text{,}\) so \(du = \frac{1}{y}dy, v = y\)):
\begin{align*} \int_1^e \frac{9}{2}\ln(y)\, dy \amp = \frac{9}{2} \Big[ y\ln(y) - y \Big]_1^e \\ \amp = \frac{9}{2} \Big( \big( e\ln(e) - e \big) - \big( 1\ln(1) - 1 \big) \Big) \\ \amp = \frac{9}{2} \Big( (e(1) - e) - (0 - 1) \Big) \\ \amp = \frac{9}{2} \Big( 0 - (-1) \Big) = \frac{9}{2} \end{align*}
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15.1.49.

  1. Which is easier, antidifferentiating \(\dfrac{y}{1 + xy}\) with respect to \(x\) or with respect to \(y\text{?}\) Explain.
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  2. Evaluate \(\ds \iint_\c{R} \dfrac{y}{1 + xy}\, dA\text{,}\) where \(\c{R} = [0,1] \times [0,1]\text{.}\)
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Solution.
  1. It is easier to antidifferentiate with respect to \(x\text{.}\) If we integrate with respect to \(x\text{,}\) the \(y\) in the numerator acts as a constant multiplier, and the expression looks like \(\int \frac{C}{1+Cx}\, dx\text{,}\) which yields a natural log function via a simple u-substitution.
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    If we try to integrate with respect to \(y\) first, we have to deal with a \(y\) in both the numerator and denominator, which would require polynomial long division (or adding/subtracting a term) to simplify before integrating.
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  2. Using Fubini’s Theorem, we will choose the easier path and integrate with respect to \(x\) first: \(dx\,dy\text{.}\)
    \begin{equation*} \int_0^1 \int_0^1 \frac{y}{1 + xy}\, dx\, dy \end{equation*}
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    Inner integral with respect to \(x\text{:}\) let \(u = 1 + xy\text{,}\) then \(du = y\, dx\text{.}\)
    \begin{align*} \int_0^1 \frac{y}{1 + xy}\, dx \amp = \Big[ \ln|1 + xy| \Big]_{x=0}^{x=1} \\ \amp = \ln|1 + y| - \ln|1 + 0| = \ln(1 + y) \end{align*}
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    Outer integral with respect to \(y\text{:}\)
    \begin{equation*} \int_0^1 \ln(1 + y)\, dy \end{equation*}
    We use Integration by Parts. Let \(u = \ln(1+y)\) and \(dv = dy\text{.}\) Then \(du = \frac{1}{1+y}dy\) and \(v = y\text{.}\) Alternatively, let \(w = 1+y\text{,}\) so \(dw = dy\text{,}\) and the bounds change from \([0,1]\) to \([1,2]\text{.}\) The integral becomes \(\int_1^2 \ln(w)\, dw\text{.}\) We know \(\int \ln(w)\, dw = w\ln(w) - w\text{.}\)
    \begin{align*} \int_1^2 \ln(w)\, dw \amp = \Big[ w\ln(w) - w \Big]_1^2 \\ \amp = (2\ln(2) - 2) - (1\ln(1) - 1) \\ \amp = 2\ln(2) - 2 - (0 - 1) \\ \amp = 2\ln(2) - 1 \end{align*}
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